Concise Mathematics Selina Solutions Class 10 ICSE Chapter 22 Heights and Distances

This chapter provides the complete ICSE Class 10 Heights and Distances Solutions from the popular Selina Concise Mathematics textbook. Heights and Distances is a fascinating application of trigonometry, where you learn to calculate the height of tall objects like towers and trees, or the width of rivers, without actually measuring them. We will use trigonometric ratios like sine, cosine, and tangent, along with the concepts of angle of elevation and angle of depression, to solve practical, real-world problems. By mastering the techniques in this chapter, you’ll be able to construct right-angled triangles from word problems and find unknown lengths and distances accurately.

If you are stuck on a particular problem or want to verify your method for solving questions on angles of elevation and depression, you have come to the right place. This page contains detailed solutions for all 60 questions from Exercise 22(A), Exercise 22(B), and the Test Yourself section. Each solution is broken down step-by-step, following the exact method prescribed by the ICSE board, ensuring you learn the correct way to present your answers in exams. Here you will find clear, accurate, and easy-to-follow solutions to master the entire chapter.

Exercise 22(A)

Question 1(a)

The measure of x is :

1. 15\sqrt{3} cm
2. 15 cm
3. 5\sqrt{3} cm
4. 5 cm

We start with the relationship given by the tangent of an angle:

Referring to the triangle △ ABC in the diagram, we have:

Substituting the given values, we have:

To find x, rearrange the equation:

To simplify, we rationalize the denominator:

This gives:

Finally, simplifying further:

Hence, Option 3 is the correct option.

Question 1(b)

The value of x is :

1. 40 m
2. 30 m
3. 20 m
4. 40\sqrt{3} m

We are given that \cos \theta = \dfrac{\text{Base}}{\text{Hypotenuse}}.

From figure,

In \triangle ABC, we have:

Substitute the known values:

Solving for x, we get:

Hence, Option 3 is the correct option.

Question 1(c)

In the given figure, AB = BD, BC = 20 cm and ∠D = 45°, the length of AC is :

1. 54.64 cm
2. 48.28 cm
3. 40 cm
4. 14.64 cm

Given that in triangle BCD, we have the angle ∠D as 45°. Applying the sine formula, we know:

sin θ = \dfrac{\text{Perpendicular}}{\text{Hypotenuse}}

In this scenario, sin 45° = \dfrac{BC}{BD}.

Substituting the values, \dfrac{1}{\sqrt{2}} = \dfrac{20}{BD}.

Solving for BD, we find BD = 20\sqrt{2} m.

According to the figure, AB is equal to BD, which gives us AB = 20\sqrt{2} m as well.

To find AC, we add AB and BC:

AC = AB + BC = 20\sqrt{2} + 20 = 28.28 + 20 = 48.28 cm.

Hence, Option 2 is the correct option.

Question 1(d)

The measure of x correct to the nearest metre is :

1. 70 m
2. 35 m
3. 61 m
4. 140 m

To find the value of x to the nearest metre, we use the trigonometric ratio for cosine:

cos θ = \dfrac{\text{Base}}{\text{Hypotenuse}}

In triangle △ ABC, we are given:

⇒ cos 30° = \dfrac{BC}{AB}

Substituting the known values, we have:

⇒ \dfrac{\sqrt{3}}{2} = \dfrac{x}{70}

Solving for x, we get:

⇒ x = \dfrac{70\sqrt{3}}{2} = 35\sqrt{3} = 60.62 \approx 61 m.

Hence, Option 3 is the correct option.

Question 1(e)

The length of DC is :

1. 10 m
2. 10(\sqrt{3} + 1) m
3. 20 m
4. 10(\sqrt{3} - 1) m

We begin by recalling the trigonometric identity:

Examining \triangle ABC, we observe:

Since \tan 45^\circ = 1, it follows that:

Solving for BC, we find:

Next, consider \triangle ABD:

Given that \tan 30^\circ = \dfrac{1}{\sqrt{3}}, we have:

Solving for BD, we get:

Finally, to find DC, we calculate:

Hence, Option 4 is the correct option.

Question 2

The height of a tree is \sqrt{3} times the length of its shadow. Find the angle of elevation of the sun.

Consider a tree represented by line segment AB, where AB is the height of the tree, and BC is the shadow cast by the tree.

Assume the shadow’s length, BC, is x meters. Consequently, the tree’s height, AB, is \sqrt{3}x meters.

Let θ be the angle of elevation of the sun. In triangle △ABC, we have:

Hence, angle of elevation of sun is 60°.

Question 3

The angle of elevation of the top of a tower from a point on the ground and at a distance of 160 m from its foot, is found to be 60°. Find the height of the tower.

Consider the tower as AB and point C, which is 160 m away from the base of the tower.

Let the height of the tower, AB, be denoted as h meters.

In the right triangle △ABC, we use the tangent of the angle of elevation:

Substituting the known values:

Since \text{tan 60°} = \sqrt{3}, we have:

Solving for h gives:

Calculating further:

Hence, the height of the tower is 277.12 m.

Question 4

A ladder is placed along a wall such that its upper end is resting against a vertical wall. The foot of the ladder is 2.4 m from the wall and the ladder is making an angle of 68° with the ground. Find the height, up to which the ladder reaches.

Consider the ladder as segment AC, where the height up to which the ladder reaches the wall is denoted as h meters.

∴ AB = h meters and BC = 2.4 meters.

Given that the ladder forms a 68° angle with the ground, we analyze △ABC.

In triangle △ABC:

⇒ \text{tan } \theta = \dfrac{\text{Perpendicular}}{\text{Base}}

⇒ \text{tan } 68° = \dfrac{AB}{BC}

⇒ 2.475 = \dfrac{h}{2.4}

⇒ h = 2.4 \times 2.475

⇒ h = 5.94 \text{ m}.

Hence, the ladder reaches up to a height of 5.94 m.

Question 5

Two persons are standing on the opposite sides of a tower. They observe the angles of elevation of the top of the tower to be 30° and 38° respectively. Find the distance between them, if the height of the tower is 50 m.

Consider the tower as PQ.

Imagine person A standing at a distance of x meters from the base of the tower (Q) and person B at a distance of y meters on the opposite side.

Given that person A observes the top of the tower at a 30° angle of elevation.

For △PQA, we have:

For person B, the angle of elevation is 38°.

In △PBQ, we get:

Thus, the total distance AB between the two persons is:

Hence, the distance between two persons = 150.6 meters.

Question 6

A boy, 1.6 m tall, is 20 m away from a tower and observes the angle of elevation of the top of the tower to be (i) 45° (ii) 60°. Find the height of the tower in each case.

(i) Consider the scenario when the angle of elevation is 45°.

In triangle △ADE:

The height of the tower, CE, is calculated as:

CE = CD + DE = 1.6 + 20 = 21.6 \text{ meters}.

Hence, height of the tower when angle of elevation is 45° is 21.6 meters.

(ii) Now, consider the case when the angle of elevation is 60°.

In triangle △ADF:

The height of the tower, CF, is:

CF = CD + DF = 1.6 + 34.64 = 36.24 \text{ meters}.

Hence, height of the tower when angle of elevation is 60° is 36.24 meters.

Question 7

The upper part of a tree, broken over by the wind, makes an angle of 45° with the ground; and the distance from the root to the point where the top of the tree touches the ground, is 15 m. What was the height of the tree before it was broken?

Consider point A as the location where the tree snaps and point C as where the top of the tree lands on the ground.

In the right triangle \triangle ABC, we employ the tangent of the angle:

Since \text{tan } 45^\circ = 1, it follows that:

Thus, AB = BC = 15 \text{ meters}.

Now, applying the Pythagorean theorem in \triangle ABC:

Substituting the known values:

Taking the square root gives:

Therefore, the total height of the tree before it was broken is:

Calculating further:

Hence, height of tree before it was broken = 36.21 meters.

Question 8

The angle of elevation of the top of an unfinished tower from a point at a distance of 80 m from its base is 30°. How much higher must the tower be raised so that its angle of elevation at the same point may be 60° ?

Consider the unfinished tower as AB and the point 80 meters away from its base as C.

From the diagram,

In △ABC,

Notice that \tan 30^\circ = \dfrac{\text{Perpendicular}}{\text{Base}}. Therefore,

Solving for AB:

Substitute BC = 80 m:

Converting to a decimal:

Thus,

Now, let’s raise the tower to point D to achieve a 60° angle of elevation.

From the diagram,

In △DBC,

\tan 60^\circ = \dfrac{\text{Perpendicular}}{\text{Base}}. So,

Calculate BD:

Substitute BC = 80 m:

Converting to a decimal:

The additional height required, AD, is:

Hence, the tower must be raised by 92.37 meters.

Question 9

At a particular time, when the sun’s altitude is 30°, the length of the shadow of a vertical tower is 45 m. Calculate :

(i) the height of the tower,

(ii) the length of the shadow of the same tower, when the sun’s altitude is :

(a) 45° (b) 60°.

(i) Assume AB represents the tower’s height.

From the diagram, in triangle △ABC, we have:

This can be expressed as:

Given that BC is the shadow of the tower, which measures 45 meters:

Solving for h, the height of the tower:

Using the approximation \sqrt{3} \approx 1.732:

Calculating gives:

Therefore, the height of the tower is 25.98 meters.

(ii) Now consider the sun’s altitude at different angles:

(a) When the sun’s altitude is 45°:

From the diagram, we have:

This implies:

Thus, the shadow length BC equals the tower’s height:

Hence, the shadow’s length when the sun’s altitude is 45° is 25.98 meters.

(b) When the sun’s altitude is 60°:

From the diagram, we have:

This gives:

Solving for BC, the shadow length:

Substitute the tower’s height:

Calculating gives:

Thus, the shadow’s length when the sun’s altitude is 60° is 15 meters.

Question 10

Two vertical poles are on either side of a road. A 30 m long ladder is placed between the two poles. When the ladder rests against one pole, it makes angle 32° 24′ with the pole and when it is turned to rest against another pole, it makes angle 32° 24′ with the road. Calculate the width of the road.

Consider two vertical poles, AP and CQ, on opposite sides of a road.

When the ladder is positioned at AB leaning against pole AP, the angle formed with the pole is ∠BAP = 32° 24′.

Refer to the diagram:

In the right triangle △ABP,

Now, when the ladder is shifted to position BC and leans against pole CQ, it forms the same angle with the road, ∠CBQ = 32° 24′.

Refer to the diagram:

In the right triangle △BQC,

Thus, the total width of the road is the sum of BP and BQ: 16.08 + 25.32 = 41.4 meters.

Therefore, the width of the road is 41.4 meters.

Question 11

Two climbers are at points A and B on a vertical cliff face. To an observer C, 40 m from the foot of the cliff, on the level ground, A is at an elevation of 48° and B of 57°. What is the distance between the climbers?

Consider point P as the base of the cliff.

From the diagram provided:

In triangle △BCP,

The tangent of angle 57° is given by:

\text{tan 57°} = \dfrac{\text{Perpendicular}}{\text{Base}} \

⇒ 1.539 = \dfrac{BP}{PC} \

⇒ BP = PC \times 1.539 \

⇒ BP = 40 \times 1.539 \

⇒ BP = 61.57 \text{ meters}.

Similarly, in triangle △ACP,

The tangent of angle 48° is given by:

\text{tan 48°} = \dfrac{\text{Perpendicular}}{\text{Base}} \

⇒ 1.110 = \dfrac{AP}{PC} \

⇒ AP = PC \times 1.110 \

⇒ AP = 40 \times 1.110 \

⇒ AP = 44.40 \text{ meters}.

The distance between the climbers, which is BA, is calculated as:

BA = BP – AP

= 61.57 – 44.40

= 17.17 meters.

Hence, distance between two climbers = 17.17 meters.

Question 12

A man stands 9 m away from a flag-pole. He observes that angle of elevation of the top of the pole is 28° and the angle of depression of the bottom of the pole is 13°. Calculate the height of pole.

Consider AC as the pole and D as the point where the man is standing.

From the diagram,

In triangle △ADB,

Now, in triangle △BDC,

Thus, the total height of the pole AC is given by:

Hence, the height of pole = 6.867 meters.

Question 13

From the top of a cliff 92 m high, the angle of depression of a buoy is 20°. Calculate, to the nearest metre, the distance of the buoy from the foot of the cliff.

Consider the cliff as line segment AB and the buoy as point C.

We know the height of the cliff, AB, is 92 m.

In the right-angled triangle △ACB, the angle of depression from the top of the cliff to the buoy is 20°.

Using the tangent ratio, we have:

This becomes:

Substituting the values, we get:

To find BC, rearrange the equation:

Calculating this gives:

Thus, the buoy is approximately 253 meters away from the foot of the cliff.

Exercise 22(B)

Question 1(a)

According to the information given in the following figure, the length of BC is :

1. \sqrt{3} m
2. (\sqrt{3} - 1) m
3. (\sqrt{3} + 1) m
4. 2\sqrt{3} m

Recall the formula for tangent in a right triangle:

Looking at \triangle ABD, we apply:

Since \tan 45^\circ = 1, we have:

This gives us BD = 1 m.

Now, consider \triangle ACD:

With \tan 30^\circ = \dfrac{1}{\sqrt{3}}, we find:

Thus, CD = \sqrt{3} m.

From the figure, the length of BC is:

Hence, Option 3 is the correct option.

Question 1(b)

The measurement of h is :

1. \sqrt{8 \times 6} cm
2. \sqrt{\dfrac{4}{3}} cm
3. (8 – 6) cm
4. 2\sqrt{2} cm

Consider the relationship given by the tangent of an angle, which is defined as:

In the context of △ ABC:

⇒ \tan 50^\circ = \dfrac{AB}{BC}

⇒ \tan 50^\circ = \dfrac{h}{6}

⇒ h = 6 \tan 50^\circ

Expressing \tan 50^\circ in terms of cotangent:

⇒ ( h = 6 \tan (90^\circ – 40^\circ) )

⇒ h = 6 \cot 40^\circ ……..(1)

Now, examine △ ABD:

⇒ \tan 40^\circ = \dfrac{AB}{BD}

⇒ \tan 40^\circ = \dfrac{h}{8}

⇒ h = 8 \tan 40^\circ ……..(2)

By multiplying equations (1) and (2), we derive:

⇒ h \times h = 6 \cot 40^\circ \times 8 \tan 40^\circ

⇒ h^2 = 8 \times 6 \times \cot 40^\circ \times \dfrac{1}{\cot 40^\circ}

⇒ h^2 = 8 \times 6

⇒ h = \sqrt{8 \times 6}.

Thus, the correct answer is option 1.

Question 1(c)

The length of AC is :

1. 22 m
2. 38 m
3. 23 m
4. 45 m

Inspecting the diagram, we see that BCDE forms a rectangle.

∴ We have BC = DE = 8 m and BE = DC = 30 m.

Recall the tangent ratio:

In the right triangle △ ABE, observe:

⇒ \tan 45° = \dfrac{AB}{BE}

⇒ 1 = \dfrac{AB}{30}

⇒ AB = 30 m.

Now, to find AC, we sum AB and BC:

AC = AB + BC = 30 + 8 = 38 m.

Hence, Option 2 is the correct option.

Question 1(d)

Using the information given in the following figure, the measurement of AE is :

1. 20 cm
2. 10 cm
3. 6\dfrac{2}{3} cm
4. 20\sqrt{3} cm

Recall the trigonometric identity:

Referring to the triangle \triangle ABC, we have:

This implies:

Solving for BC, we find:

Considering the rectangle BCDE, the opposite sides are equal, thus:

Now, examining \triangle AED, we use:

Which gives us:

Solving for AE, we get:

Hence, Option 3 is the correct option.

Question 1(e)

The length of DC is :

1. 15\sqrt{3} cm
2. 15(\sqrt{3} - 1) cm
3. 15(\sqrt{3} + 1) cm
4. 10\sqrt{3} cm

Consider the formula for the tangent of an angle:

In \triangle ABC, observe that:

Substitute the known values:

Solving for BC, we have:

To eliminate the square root in the denominator, multiply by \dfrac{\sqrt{3}}{\sqrt{3}}:

Simplify the expression:

Which further simplifies to:

Now, in \triangle ABD:

Substitute the values:

Solving for BD, we find:

From the figure, the length of DC can be calculated as:

Hence, Option 4 is the correct option.

Question 2

In the figure, given below, it is given that AB is perpendicular to BD and is of length X metres. DC = 30 m, ∠ADB = 30° and ∠ACB = 45°. Find X.

Consider the given figure.

For △ABD, we apply the tangent ratio:

This gives us:

Solving for AB, we have:

Next, examine △ABC:

This simplifies to:

Thus, we find:

By equating the expressions from (1) and (2), we get:

Substituting for BD, we have:

Since BD = BC + CD, it follows:

Rearranging terms gives:

Factoring out BC, we obtain:

Solving for BC, we find:

Substituting the approximate value of \sqrt{3}:

Calculating gives:

Therefore, AB = X = 40.98 meters, as per equation (2).

Hence, X = 40.98 meters.

Question 3

Find the height of a tree when it is found that on walking away from it 20 m, in a horizontal line through its base, the elevation of its top changes from 60° to 30°.

Consider the tree as segment AB.

Let points C and D be such that the distance between them, CD, is 20 m. The angle of elevation ∠ACB is 60°, and ∠ADB is 30°.

In triangle ∆ABC, observe that:

This gives us:

Thus, we find:

Now, in triangle ∆ABD, note:

This results in:

Hence, we have:

By equating equations (1) and (2), we derive:

This implies:

Given that:

We find:

Thus, solving for BC, we have:

Substituting the value of CD:

Therefore:

Finally, calculating AB:

Hence, the height of the tree is 17.32 metres.

Question 4

From the top of a light house 100 m high, the angles of depression of two ships are observed as 48° and 36° respectively. Find the distance between the two ships (in the nearest metre) if:

(i) the ships are on the same side of the light house.

(ii) the ships are on the opposite sides of the light house.

(i) Consider the lighthouse as AB.

The angles of depression are given as 48° and 36°.

When both ships are on the same side of the lighthouse:

In triangle ABC,

In triangle ABD,

Thus, the distance between the two ships (CD) is calculated as:

CD = BD – BC = 137.64 – 90.04 = 47.6 \approx 48 \text{ m}.

Hence, distance between ships when on the same side = 48 m.

(ii) Assume the lighthouse as AB again.

The angles of depression are 48° and 36°.

Since alternate angles are equal, we have:

∠ADB = ∠QAD = 36° and ∠ACB = ∠PAC = 48°.

When the ships are on opposite sides:

In triangle ABC,

In triangle ABD,

Therefore, the distance between the two ships (CD) is:

CD = BD + BC = 137.64 + 90.04 = 227.68 \approx 228 \text{ m}.

Hence, the distance between two ships, when on opposite side = 228 m.

Question 5

Two pillars of equal heights stand on either side of a roadway, which is 150 m wide. At a point in the roadway between the pillars the elevations of the tops of the pillars are 60° and 30°; find the height of the pillars and the position of the point.

Consider two towers, AB and CD, each with a height of h meters. A point P lies on the road BD, which is 150 meters wide. The angles of elevation from P to the tops of the towers are 60° and 30° respectively.

In the triangle ABP, we apply the tangent of 60°:

Similarly, in triangle CDP, using the tangent of 30°:

Since BD is 150 meters, we have:

Substituting from equations (1) and (2):

\Rightarrow h = 64.95 \text{ meters}.

Using equation (1), we find:

BP = \dfrac{h}{\sqrt{3}} = \dfrac{64.95}{1.732} = 37.5 \text{ meters}.

Thus, each pillar has a height of 64.95 meters, and point P is located 37.5 meters from pillar AB.

Question 6

From the figure, given below, calculate the length of CD.

Observe the figure provided.

We have DE = CB = 15 m.

Consider the triangle ∆AED:

The tangent of angle 22° is given by:

Substituting the known values, we find:

Solving for AE, we multiply both sides by 15:

Now, consider the triangle ∆ABC:

The tangent of angle 47° is:

Substituting the known values, we have:

To find AB, multiply both sides by 15:

From the figure, we understand:

CD = BE = AB – AE

Substituting the values we calculated:

= 16.086 – 6.06

= 10.03 meters.

Hence, CD = 10.03 meters.

Question 7

The angle of elevation of the top of a tower is observed to be 60°. At a point, 30 m vertically above the first point of observation, the elevation is found to be 45°. Find:

(i) the height of the tower,

(ii) its horizontal distance from the points of observation.

(i) Consider the tower as AB, with C as the initial observation point and D as the second point.

From the diagram,

Let BC = ED = a, and BE = CD = 30 m.

In triangle △ABC, we have:

In triangle △AED, observe:

Since CD = AB – AE, we have:

Simplifying gives:

Thus:

Using equation (1), we find:

Therefore, the height of the tower is 70.98 meters.

(ii) From our previous calculation, we know:

ED = a = 40.98 meters.

Thus, the horizontal distance from the points of observation is 40.98 meters.

Question 8

From the top of a cliff, 60 meters high, the angles of depression of the top and bottom of a tower are observed to be 30° and 60°. Find the height of the tower.

Consider the cliff as CD, which measures 60 meters, and the tower as AB.

Since alternate interior angles are equal, we have:

∴ ∠ECA = ∠CAF = 30° and ∠ECB = ∠CBD = 60°.

Let AF = BD = a meters.

In △BCD:

In △AFC:

From the diagram, we find:

⇒ AB = DF = CD – CF

⇒ AB = CD – CF

⇒ AB = 60 – 20 = 40 meters.

Hence, the height of the tower = 40 meters.

Question 9

A man on a cliff observes a boat, at an angle of depression 30°, which is sailing towards the shore to the point immediately beneath him. Three minutes later, the angle of depression of the boat is found to be 60°. Assuming that the boat sails at a uniform speed, determine :

(i) how much more time it will take to reach the shore ?

(ii) the speed of the boat in metre per second, if the height of the cliff is 500 m.

Consider the cliff as CD, where the boat was initially at point A with a 30° angle of depression, and later at point B with a 60° angle of depression.

(i) Observing from the figure, we have the following:

For the angle of 60°:

For the angle of 30°:

Dividing equation (1) by equation (2), we derive:

The boat moves from point A to B in 3 minutes.

Thus, the distance AB is covered in 3 minutes, meaning:

⇒ 2 BD in 3 minutes

⇒ BD in \dfrac{3}{2} = 1.5 minutes.

Therefore, it takes an additional 1.5 minutes to reach the shore.

(ii) In triangle △ADC:

In triangle △BDC:

From the figure, we know:

Assume the speed of the boat is a meters per second, then in 3 minutes the boat travels:

Distance (AB) = Speed × Time

AB = a × 3 × 60

AB = 180a meters ………(2)

Equating (1) and (2), we get:

Therefore, the speed of the boat is 3.21 m/s.

Question 10

A man in a boat rowing away from a lighthouse 150 m high, takes 2 minutes to change the angle of elevation of the top of the lighthouse from 60° to 45°. Find the speed of the boat.

Consider the initial position of the man in the boat as point C and after 2 minutes, he reaches point D. The lighthouse is represented by AB.

AB = 150 meters.

In triangle △ABC,

\text{tan 60°} = \frac{\text{Perpendicular}}{\text{Base}}
\Rightarrow \sqrt{3} = \frac{AB}{BC}
\Rightarrow AB = \sqrt{3} \times BC
\Rightarrow BC = \frac{AB}{\sqrt{3}}
\Rightarrow BC = \frac{150}{\sqrt{3}}
\Rightarrow BC = \frac{150}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}
\Rightarrow BC = \frac{150\sqrt{3}}{3}
\Rightarrow BC = 50\sqrt{3} \text{ meters}.

Now, in triangle △ABD,

\text{tan 45°} = \frac{\text{Perpendicular}}{\text{Base}}
\Rightarrow 1 = \frac{AB}{BD}
\Rightarrow AB = BD
\Rightarrow BD = 150 \text{ metres}.

The distance covered by the boat, CD, is given by:

CD = BD – BC = 150 – 50\sqrt{3}

= 150 – 86.6

= 63.4 meters.

The boat travels 63.4 meters in 2 minutes, or 120 seconds.

The speed of the boat is calculated as:

Hence, the speed of boat = 0.53 m/sec.

Question 11

A person standing on the bank of a river observes that the angle of elevation of the top of a tree standing on the opposite bank is 60°. When he moves 40 m away from the bank, he finds the angle of elevation to be 30°. Find :

(i) the height of the tree, correct to 2 decimal places.

(ii) the width of the river.

Assume the tree is represented by CD, with B as the initial position where the angle of elevation is 60°, and A as the position after moving 40 m away, where the angle of elevation is 30°.

(i) Consider △BCD:

The tangent of 60° is given by:

This leads to:

Thus, we find:

Now, consider △ACD:

The tangent of 30° is:

Which gives:

So, we have:

From the diagram, notice that:

AB = AC – BC

Therefore:

Simplifying gives:

Which results in:

Solving for CD, we find:

Finally, calculating the numerical value:

Thus, the height of the tree is 34.64 meters.

(ii) From the previous calculation, we determined:

This simplifies to:

20 meters.

Therefore, the width of the river is 20 meters.

Question 12

The horizontal distance between two towers is 75 m and the angular depression of the top of the first tower as seen from the top of the second, which is 160 m high, is 45°. Find the height of the first tower.

Consider two towers, where AB represents the first tower and CD represents the second tower. The angle of depression from the top of the second tower to the top of the first tower is ∠EDA = 45°.

Since the angle of depression equals the angle of elevation from the horizontal, we have:

∴ ∠DAF = ∠EDA = 45°.

The horizontal distance between the towers is given as AF = BC = 75 m.

In triangle △ADF, applying the tangent of 45° gives:

Now, to find the height of the first tower AB, we use the relation:

AB = FC = CD – DF = 160 – 75 = 85 \text{ m}.

Therefore, the height of the first tower is 85 m.

Question 13

The length of the shadow of a tower standing on level plane is found to be 2y meters longer when the sun’s altitude is 30° than when it was 45°. Prove that the height of the tower is y(\sqrt{3} + 1) meters.

Consider a tower CD with height h meters. Let BC be the shadow length when the sun’s elevation is 45°, and AC be the shadow length when the sun’s elevation is 30°.

In the triangle △ACD, we use the tangent of 30°:

Now, in triangle △BCD, the tangent of 45° is applied:

The length AC can be expressed as the sum of AB and BC:

AC = AB + BC = (2y + h) \text{ meters}.

Substituting the expression for AC into equation (1), we have:

\begin{aligned}\Rightarrow h = \dfrac{2y + h}{\sqrt{3}} \\\Rightarrow \sqrt{3}h = 2y + h \\\Rightarrow \sqrt{3}h - h = 2y \\\Rightarrow h(\sqrt{3} - 1) = 2y \\\Rightarrow h = \dfrac{2y}{\sqrt{3} - 1}\end{aligned}.

To simplify, multiply the numerator and denominator by (\sqrt{3} + 1):

\begin{aligned}\Rightarrow h = \dfrac{2y}{\sqrt{3} - 1} \times \dfrac{\sqrt{3} + 1}{\sqrt{3} + 1} \\\Rightarrow h = \dfrac{2y(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} \\\Rightarrow h = \dfrac{2y(\sqrt{3} + 1)}{(\sqrt{3})^2 - 1^2} \\\Rightarrow h = \dfrac{2y(\sqrt{3} + 1)}{3 - 1} \\\Rightarrow h = \dfrac{2y(\sqrt{3} + 1)}{2} \\\Rightarrow h = y(\sqrt{3} + 1)\end{aligned}.

Thus, we have shown that the height of the tower is y(\sqrt{3} + 1) meters.

Question 14

An aeroplane flying horizontally 1 km above the ground and going away from the observer is observed at an elevation of 60°. After 10 seconds, its elevation is observed to be 30°; find the uniform speed of the aeroplane in km per hour.

Imagine the aeroplane initially at point E, and after 10 seconds, it moves to point D.

In triangle △ABE, notice:

\text{tan 60°} = \dfrac{\text{Perpendicular}}{\text{Base}}
⇒ \sqrt{3} = \dfrac{BE}{AB}
⇒ \sqrt{3} = \dfrac{1}{AB}
⇒ AB = \dfrac{1}{\sqrt{3}} \text{ km}.

Now, in triangle △ACD:

\text{tan 30°} = \dfrac{\text{Perpendicular}}{\text{Base}}
⇒ \dfrac{1}{\sqrt{3}} = \dfrac{CD}{AC}
⇒ AC = \sqrt{3}CD
⇒ AC = \sqrt{3} \times 1 = \sqrt{3} \text{ km}.

From the diagram, we see:

DE = BC.

To find BC, calculate:

BC = AC - AB = \sqrt{3} - \dfrac{1}{\sqrt{3}}
= \dfrac{3 - 1}{\sqrt{3}}
= \dfrac{2}{1.732}
= 1.1547 \text{ km}.

∴ DE = 1.1547 km.

Thus, the aeroplane covers 1.1547 km in 10 seconds.

Convert time: 10 seconds = \dfrac{10}{3600} = \dfrac{1}{360} hours.

Now, calculate speed:

\text{Speed} = \dfrac{\text{Distance}}{\text{Time}} = \dfrac{1.1547}{\dfrac{1}{360}}
= 1.1547 \times 360
= 415.69 \text{ km/hr}.

Hence, speed of aeroplane = 415.69 km/hr.

Question 15

From the top of a hill, the angles of depression of two consecutive kilometer stones, due east are found to be 30° and 45° respectively. Find the distances of the two stones from the foot of hill.

Consider the positions of the kilometer stones as C and D, and the hill as AB.

In the right triangle △ABD, the angle of depression to stone D is 30°. Therefore, we have:

Similarly, in △ABC, where the angle of depression to stone C is 45°:

From the geometric arrangement, it follows that:

⇒ CD = BD – BC

Substituting the expressions from equations (1) and (2), we have:

⇒ 1 = \sqrt{3}AB - AB

⇒ 1 = AB(\sqrt{3} - 1)

Solving for AB, we get:

⇒ AB = \dfrac{1}{\sqrt{3} - 1} = \dfrac{1}{0.732} = 1.366 km.

From equation (2), since BC = AB, it follows:

BC = 1.366 km

For BD, we use:

BD = BC + CD = 1.366 + 1 = 2.366 km.

Hence, kilometer stones are at a distance of 1.366 and 2.366 km.

Test Yourself

Question 1(a)

If CD = 10 m, the length of AB is :

1. \dfrac{40}{\sqrt{3}} m
2. \dfrac{20}{\sqrt{3}} m
3. 100 m
4. 30 m

Consider the given problem, where we need to find the length of AB.

Notice that in triangle BCD, we apply the tangent ratio:

Given \tan 30^\circ = \dfrac{1}{\sqrt{3}}, and CD = 10 m, we have:

Solving for BD gives:

Next, in triangle ACD, we again use the tangent ratio:

With \tan 60^\circ = \sqrt{3}, and CD = 10 m, we find:

Solving for AD yields:

Now, to find AB, we add AD and BD:

To simplify:

Hence, Option 1 is the correct option.

Question 1(b)

If tan x° = \dfrac{5}{2}, the length of CB is :

1. 0.4 m
2. 40 m
3. 50 m
4. 80 m

Recall the identity for tangent in a right triangle:

In the given triangle \triangle ABC, we have:

Given that \tan x^\circ = \dfrac{5}{2}, we can equate:

To find CB, cross-multiply to get:

Hence, Option 2 is the correct option.

Question 1(c)

The length of AC is :

1. (60 - 20\sqrt{3}) m
2. 60\sqrt{3} m
3. (60 + 20\sqrt{3}) m
4. 20\sqrt{3} m

Consider the rectangle BCDE where opposite sides are of equal length. Thus, BE equals DC, both measuring 60 m.

Recall the tangent formula for a right triangle:

Now, look at triangle ABE:

• For angle 30°, \tan 30° = \dfrac{AB}{BE}.
• Substitute the values: \dfrac{1}{\sqrt{3}} = \dfrac{AB}{60}.
• Solving for AB gives AB = \dfrac{60}{\sqrt{3}} = 20\sqrt{3} \text{ m}.

Next, examine triangle EBC:

• For angle 45°, \tan 45° = \dfrac{BC}{BE}.
• Here, 1 = \dfrac{BC}{60}.
• This implies BC is 60 m.

Therefore, the length of AC is the sum of AB and BC:

Hence, Option 3 is the correct option.

Question 1(d)

BCDE is a square with side 90 cm and ∠F = 45°. The length of AF is :

1. 60\sqrt{2} cm
2. 90\sqrt{2} cm
3. 120\sqrt{2} cm
4. 180\sqrt{2} cm

Given that BCDE is a square, we have:

∴ BE = CD = 90 cm.

Recall the trigonometric identity:

From the diagram, notice:

∠AEB = ∠EFD = 45° (since corresponding angles are equal)

Considering △ABE,

⇒ \tan 45° = \dfrac{AB}{BE}

⇒ 1 = \dfrac{AB}{90}

⇒ AB = 90 cm.

From the diagram:

AC = AB + BC = 90 + 90 = 180 cm.

Recall another trigonometric identity:

In △ACF,

⇒ \sin 45° = \dfrac{AC}{AF}

⇒ \dfrac{1}{\sqrt{2}} = \dfrac{180}{AF}

⇒ AF = 180\sqrt{2} cm.

Hence, Option 4 is the correct option.

Question 1(e)

The length of DC is :

1. 45\sqrt{3} m
2. (45 - \sqrt{3}) m
3. 45(\sqrt{3} - 1) m
4. 45(\sqrt{3} + 1) m

We have the formula for tangent as:

Looking at the triangle ABC:

In △ ABC, we have:

⇒ \tan 45° = \dfrac{AB}{BC}

Since \tan 45° = 1, this gives us:

⇒ 1 = \dfrac{AB}{45}

⇒ AB = 45 m.

Now, considering triangle ABD:

In △ ABD, we have:

⇒ \tan 30° = \dfrac{AB}{BD}

Given \tan 30° = \dfrac{1}{\sqrt{3}}, we find:

⇒ \dfrac{1}{\sqrt{3}} = \dfrac{45}{BD}

⇒ BD = 45\sqrt{3} m.

From the figure, we can see:

DC = BD - BC = 45\sqrt{3} - 45 = 45(\sqrt{3} - 1) m.

Hence, Option 3 is the correct option.

Question 1(f)

The ratio of the length of a vertical pole and length of its shadow on the horizontal surface is 3 : \sqrt{3}

Assertion(A): The angle of elevation of the sun is 60°.

Reason(R): If angle of elevation of the sun is θ, tan θ = \dfrac{3}{\sqrt{3}} = \sqrt{3} = tan 60°.

• (a) A is true, R is false.
• (b) A is false, R is true.
• (c) Both A and R are true and R is correct reason for A.
• (d) Both A and R are true and R is incorrect reason for A.

We’re given that the vertical pole’s height and its shadow’s length have a ratio of 3 : \sqrt{3}.

Consider AB as the pole and BC as the shadow, with θ as the angle of elevation of the sun.

⇒ \dfrac{\text{Height of the pole}}{\text{Length of the shadow}} = \dfrac{3}{\sqrt{3}}

⇒ \dfrac{\text{AB}}{\text{BC}} = \sqrt{3}

⇒ \text{tan } θ = \sqrt{3}

⇒ \text{tan } θ = \text{tan } 60°

⇒ θ = 60°

∴ Both A and R are true, and R provides the correct reasoning for A.

Hence, option 3 is the correct option.

Question 1(g)

The length of the ladder placed against a vertical wall is twice the distance between the foot of the ladder and the wall.

Assertion(A): The angle that the ladder makes with the wall is 60°.

Reason(R): If ladder makes angle θ with the wall then sin θ = \dfrac{\text{base}}{\text{hypotenuse}} = \dfrac{x}{2x} = \dfrac{1}{2}.

• (a) A is true, R is false.
• (b) A is false, R is true.
• (c) Both A and R are true and R is correct reason for A.
• (d) Both A and R are true and R is incorrect reason for A.

Consider a ladder placed against a vertical wall. Let AB represent the wall and AC the ladder. Suppose the distance between the foot of the ladder (C) and the wall (B) is x.

Thus, the length of the ladder AC is 2x.

Applying the Pythagorean theorem:

∴ Hypotenuse² = Base² + Height²

⇒ AC² = BC² + AB²

⇒ (2x)² = x² + AB²

⇒ 4x² = x² + AB²

⇒ AB² = 4x² – x²

⇒ AB² = 3x²

⇒ AB = \sqrt{3x^2}

⇒ AB = \sqrt{3}x.

Now, let θ be the angle between the ladder and the wall.

We know:

⇒ \text{sin } θ = \dfrac{\text{Perpendicular side}}{\text{Hypotenuse}}

⇒ \text{sin } θ = \dfrac{BC}{AC}

⇒ \text{sin } θ = \dfrac{x}{2x}

⇒ \text{sin } θ = \dfrac{1}{2}

This implies:

⇒ \text{sin } θ = \text{sin } 30°

⇒ θ = 30°.

Thus, the assertion (A) that the angle is 60° is incorrect.

However, the reasoning (R) given is true, as it correctly calculates \text{sin } θ = \dfrac{x}{2x} = \dfrac{1}{2}, indicating θ is 30°.

Hence, option 2 is the correct option.

Question 1(h)

For the following figure, tan A = 1\dfrac{2}{3} and tan B = \dfrac{1}{3}

Statement (1): x = 36 cm

Statement (2): tan A = \dfrac{5}{3} = \dfrac{h}{9} ⇒ h = 15

tan B = \dfrac{h}{x + 9} \Rightarrow \dfrac{1}{3} = \dfrac{15}{x + 9}

• (a) Both the statements are true.
• (b) Both the statements are false.
• (c) Statement 1 is true, and statement 2 is false.
• (d) Statement 1 is false, and statement 2 is true.

We have the values for tan A as 1\dfrac{2}{3} = \dfrac{5}{3} and tan B as \dfrac{1}{3}.

From the diagram, we can say:

This implies:

By cross-multiplying, we find:

Now, considering tan B:

Substitute the values:

Cross-multiplying gives:

Subtracting 9 from both sides, we get:

∴ Both the statements are indeed true.

Hence, option 1 is the correct option.

Question 1(i)

CD = 100 m, ∠ADB = 15° and ∠BDC = 45°

Statement (1): AB = \dfrac{\text{tan 60° - tan 45°}}{100}

Statement (2): AB = (100 tan 60° – 100 tan 45°) m.

• (a) Both the statements are true.
• (b) Both the statements are false.
• (c) Statement 1 is true, and statement 2 is false.
• (d) Statement 1 is false, and statement 2 is true.

Consider the given figure.

In \triangle BDC:

\Rightarrow \text{tan} 45° = \dfrac{BC}{DC}$$1em]
\Rightarrow \text{tan} 45° = \dfrac{BC}{100}[1em]
\Rightarrow BC = 100 \times \text{tan} 45°[1em]

Now, in \triangle ADC:

\Rightarrow \text{tan} 60° = \dfrac{AC}{DC}[1em]
\Rightarrow \text{tan} 60° = \dfrac{AB + BC}{100}[1em]
\Rightarrow \text{tan} 60° = \dfrac{AB + 100 \times \text{tan} 45°}{100}[1em]
\Rightarrow 100 \times \text{tan} 60° = AB + 100 \times \text{tan} 45°[1em]
\Rightarrow AB = 100 \times \text{tan} 60° - 100 \times \text{tan} 45°

∴ Statement 1 is false, and statement 2 is true.

Hence, option 4 is the correct option.

Question 2

Find AD.

(i)

(ii)

(i) Let’s examine the figure.

We have that BE = CD = 20 m and DE = CB = 5 m.

In the triangle ABE, we apply the tangent function:

To find AD, add AE and DE:

AD = AE + DE = 12.498 + 5 = 17.498 \approx 17.5 \text{ m}.

Thus, AD = 17.5 meters.

(ii) Let’s use the exterior angle theorem:

An exterior angle is the sum of the two opposite interior angles.

∴ ∠ACD = ∠ABC + ∠BAC

Since ∠ABC = ∠BAC (because angles opposite equal sides are equal),

∴ ∠ACD = 2∠ABC

⇒ 2∠ABC = 48°

⇒ ∠ABC = 24°.

In triangle ABD, we use the sine function:

Thus, AD = 12.20 meters.

Question 3

In the following diagram, AB is a floor-board; PQRS is a cubical box with each edge = 1 m and ∠B = 60°. Calculate the length of the board AB.

In the triangle \Delta PSB, notice that we can use the sine function for angle 60^\circ:

\text{sin 60°} = \dfrac{\text{Perpendicular}}{\text{Hypotenuse}}.

This gives us:

\Rightarrow \dfrac{\sqrt{3}}{2} = \dfrac{PS}{PB}.

Solving for PB, we have:

\Rightarrow PB = \dfrac{2PS}{\sqrt{3}}.

Since PS = 1 m, substituting this value in gives:

\Rightarrow PB = \dfrac{2 \times 1}{1.732} = 1.155 \text{ m}.

Now, consider \Delta APQ. Here, the angle \angle APQ is equivalent to \angle ABR, both being 60^\circ due to corresponding angles.

Using the cosine function:

\text{cos 60°} = \dfrac{\text{Base}}{\text{Hypotenuse}}.

This results in:

\Rightarrow \dfrac{1}{2} = \dfrac{PQ}{AP}.

Solving for AP, we find:

\Rightarrow AP = 2PQ.

Given PQ = 1 m, we substitute to obtain:

\Rightarrow AP = 2 \text{ m}.

From the diagram, the total length of the board AB is the sum of AP and PB:

AB = AP + PB = 2 + 1.155 = 3.155 \text{ m}.

Hence, AB = 3.155 meters.

Question 4

Calculate BC.

Consider the triangle ∆ABD,

[ \text{tan 35°} = \dfrac{\text{Perpendicular}}{\text{Base}}$$

This implies:

Thus, we can express BD as:

Substituting the value of AD:

Calculating further, we find:

Now, examine the triangle ∆ACD,

This gives us:

So, CD can be written as:

Substituting the known value of AD:

After calculating, we obtain:

According to the figure, the length of BC is determined by:

Therefore, BC = 10.55 meters.

Question 5

Calculate AB.

Consider △ACD.

Using the cosine of 30°, we have:

This gives us:

Solving for AD, we find:

Substituting the value of CD as 6:

Now, examining △BDE:

The sine of 47° is expressed as:

Thus, we have:

This implies:

Substitute DE = 5:

From the diagram, we see:

Rounding off gives us:

Therefore, AB = 8.85 meters.

Question 6

The radius of a circle is given as 15 cm and chord AB subtends an angle of 131° at the centre C of the circle. Using trigonometry, calculate :

(i) the length of AB;

(ii) the distance of AB from the centre C.

We have the circle with center C and radius CA = CB = 15 cm, and the angle ∠ACB = 131°.

To solve for the chord AB, let’s draw a perpendicular line CP from the center C to the chord AB.

Notice that this perpendicular bisects the chord AB, so CP divides AB into two equal parts.

In triangles △ACP and △BCP:

• ∠APC = ∠BPC = 90° (because CP is perpendicular to AB)
• CP is common in both triangles
• AP = PB (since CP bisects AB)

Therefore, by the SAS axiom, △ACP ≅ △BCP.

This implies ∠ACP = ∠BCP = \dfrac{131°}{2} = 65.5° by C.P.C.T. (Corresponding Parts of Congruent Triangles).

Consider △ACP, we use the sine function:

(i) From the figure, the length of AB is:

Hence, AB = 27.30 cm.

(ii) Now, to find the distance of AB from the center C, in △ACP:

We use the cosine function:

Hence, CP = 6.225 cm.

Question 7

At a point on level ground, the angle of elevation of a vertical tower is found to be such that its tangent is \dfrac{5}{12}. On walking 192 meters towards the tower; the tangent of the angle is found to be \dfrac{3}{4}. Find the height of the tower.

Assume the vertical tower is AB, and the points on the ground are C and D such that the distance between C and D is 192 m.

Let ∠ACB = θ and ∠ADB = α.

Given that:

∴ \text{tan } θ = \dfrac{5}{12}

⇒ \dfrac{\text{Perpendicular}}{\text{Base}} = \dfrac{5}{12}

⇒ \dfrac{AB}{BC} = \dfrac{5}{12}

⇒ AB = \dfrac{5}{12}BC ………(1)

Also given:

∴ \text{tan } α = \dfrac{3}{4}

⇒ \dfrac{\text{Perpendicular}}{\text{Base}} = \dfrac{3}{4}

⇒ \dfrac{AB}{BD} = \dfrac{3}{4}

⇒ AB = \dfrac{3}{4}BD ………(2)

From equations (1) and (2), we have:

∴ \dfrac{5}{12}BC = \dfrac{3}{4}BD

⇒ \dfrac{BC}{BD} = \dfrac{3 \times 12}{4 \times 5}

⇒ \dfrac{BD + CD}{BD} = \dfrac{36}{20}

⇒ \dfrac{BD + CD}{BD} = \dfrac{9}{5}

⇒ 5(BD + CD) = 9BD

⇒ 5BD + 5CD = 9BD

⇒ 9BD - 5BD = 5CD

⇒ 4BD = 5CD

⇒ BD = \dfrac{5 \times 192}{4}

⇒ BD = \dfrac{960}{4}

⇒ BD = 240 \text{ m}.

Now, BC = BD + DC = 240 + 192 = 432 \text{ m}.

Using equation (1),

= 5 × 36

= 180 m.

Hence, the height of the tower is 180 m.

Question 8

A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff of height h meter. At a point on the plane, the angle of elevation of the bottom of the flagstaff is α and at the top of the flagstaff is β. Prove that the height of the tower is \dfrac{\text{h tan α}}{\text{tan β} - \text{tan α}}.

Consider a vertical tower AB with height x meters, topped by a flagstaff AD of height h meters. Point C is located on the horizontal plane such that the angle of elevation to the bottom of the flagstaff at B is α, and to the top at D is β.

In the triangle ABC, we have:

This implies:

Rearranging gives:

Thus:

Now, consider triangle DBC:

So:

Substitute the expression for BC from equation (1):

Now express BD as the sum of BA and AD:

Multiply through by \dfrac{x}{\text{tan } \alpha}:

Rearrange to get:

Distribute \text{tan } \alpha:

Rearrange terms:

Factor out x:

Finally, solve for x:

Thus, the height of the tower is proved to be \dfrac{h \text{tan } \alpha}{\text{tan } \beta - \text{tan } \alpha}.

Question 9

With reference to the given figure, a man stands on the ground at point A, which is on the same horizontal plane as B, the foot of the vertical pole BC. The height of the pole is 10 m. The man’s eye is 2 m above the ground. He observes the angle of elevation of C, the top of the pole, as x°, where tan x° = \dfrac{2}{5}. Calculate:

(i) the distance AB in metres;

(ii) angle of elevation of the top of the pole when he is standing 15 metres from the pole. Give your answer to the nearest degree.

Consider the height of the man’s eyes from the ground as AD, which is 2 m.

Notice from the figure that BE is equal to AD, thus BE = 2 m.

Now, calculate CE:

CE = BC – BE = 10 – 2 = 8 m.

(i) In the triangle CED, apply the tangent ratio:

This gives us:

Substitute the given values:

Solving for DE:

From the figure, AB equals DE, therefore:

Hence, AB = 20 m.

(ii) Let A’D’ represent the man’s new position with θ as the angle of elevation to the top of the pole.

Given that D’E = 15 m, in triangle CED’, the tangent ratio is:

Therefore:

Substitute the known values:

This corresponds to:

Thus:

Hence, the angle of elevation of the top of the pole when the man is standing 15 metres from the pole is 28°.

Question 10

From a window A, 10 m above the ground the angle of elevation of the top C of a tower is x°, where tan x° = \dfrac{5}{2} and the angle of depression of the foot D of the tower is y°, where tan y° = \dfrac{1}{4}. Calculate the height CD of the tower in metres.

Consider the given figure.

Notice that AB = DE = 10 m.

Let’s analyze ∆AED:

\text{tan } y° = \frac{\text{Perpendicular}}{\text{Base}}
\Rightarrow \text{tan } y° = \frac{DE}{AE}
\Rightarrow \frac{1}{4} = \frac{DE}{AE}
\Rightarrow AE = 4 \times DE = 4 \times 10 = 40 \text{ m}.

Now, consider ∆AEC:

\text{tan } x° = \frac{\text{Perpendicular}}{\text{Base}}
\Rightarrow \text{tan } x° = \frac{CE}{AE}
\Rightarrow \frac{5}{2} = \frac{CE}{AE}
\Rightarrow CE = AE \times \frac{5}{2} = 40 \times \frac{5}{2} = 100 \text{ m}.

From the figure, the total height of the tower CD is given by:

CD = DE + CE = 10 + 100 = 110 m.

Hence, height of tower (CD) = 110 m.

Question 11

A vertical tower is 20 m high. A man standing at some distance from the tower knows that the cosine of the angle of elevation of the top of the tower is 0.53. How far is he standing from the foot of the tower ?

Let’s assume the angle of elevation is denoted by \theta.

According to the given information, we have:

⇒ \cos \theta = 0.53

This implies:

⇒ \cos \theta = \cos 58^\circ

Thus, \theta = 58^\circ.

To find \sin \theta, use the identity:

⇒ \cos^2 \theta = 0.2809

⇒ 1 - \sin^2 \theta = 0.2809

⇒ \sin^2 \theta = 1 - 0.2809

⇒ \sin^2 \theta = 0.7191

Taking the square root on both sides gives:

⇒ \sin \theta = \sqrt{0.7191}

⇒ \sin \theta = 0.848

Now, calculate \tan \theta using \sin \theta and \cos \theta:

⇒ \tan \theta = \dfrac{\text{sin } \theta}{\text{cos } \theta} = \dfrac{0.848}{0.53} = 1.6

This results in the ratio:

The man is standing at a distance of 12.5 meters.

Question 12

A vertical pole and a vertical tower are on the same level ground in such a way that from the top of the pole the angle of elevation of the top of the tower is 60° and the angle of depression of the bottom of the tower is 30°. Find :

(i) the height of the tower, if the height of the pole is 20 m;

(ii) the height of the pole, if the height of the tower is 75 m

(i) Consider AB as the pole and CD as the tower.

Given that the height of the pole, AB, is 20 m.

From the diagram, we have CE = AB = 20 m.

In triangle AEC, use the tangent of 30°:

Now, in triangle AED, use the tangent of 60°:

According to the diagram, CD = DE + CE:

Therefore, the height of the tower is 80 m.

(ii) Now, if the height of the tower, CD, is 75 m, let AE = BC = y.

In triangle AEC, using the tangent of 30°:

In triangle AED, using the tangent of 60°:

From the diagram, CD = CE + DE:

We find CE as:

Hence, the height of the pole is 18.75 meters.

Question 13

In the given figure, from the top of a building AB = 60 m high, the angles of depression of the top and bottom of a vertical lamp post CD are observed to be 30° and 60° respectively. Find :

(i) the horizontal distance between AB and CD.

(ii) the height of the lamp post.

(i) Notice that alternate angles are equal, so we have:

∴ ∠ACB = ∠EAC = 60°

In △ABC, apply the tangent ratio:

\text{tan 60°} = \dfrac{\text{Perpendicular}}{\text{Base}}
\Rightarrow \sqrt{3} = \dfrac{AB}{BC}
\Rightarrow BC = \dfrac{AB}{\sqrt{3}} = \dfrac{60}{1.732} = 34.64 \text{ m}.

Thus, the horizontal distance between AB and CD is 34.64 meters.

(ii) From the figure, note that:

FD = BC = 34.64 m

Again, using alternate angles, we know:

∴ ∠ADF = ∠EAD = 30°

In △AFD, use the tangent ratio:

\text{tan 30°} = \dfrac{\text{Perpendicular}}{\text{Base}}
\Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{AF}{FD}
\Rightarrow AF = \dfrac{FD}{\sqrt{3}}
\Rightarrow AF = \dfrac{34.64}{1.732} = 20 \text{ m}.

From the figure, calculate BF:

BF = AB – AF = 60 – 20 = 40 m.

∴ CD = 40 m.

Therefore, the height of the lamp post is 40 meters.

Question 14

An aeroplane, at an altitude of 250 m, observes the angle of depression of two boats on the opposite banks of a river to be 45° and 60° respectively. If the boats are on the opposite sides of the aeroplane, find the width of the river. Write the answer correct to the nearest whole number.

Consider the aeroplane at point D and the boats at points A and B on opposite banks of the river.

Given the angles of depression are 60° and 45°.

Since alternate interior angles are equal in this scenario:

∴ ∠DAC = EDA = 60° and ∠DBC = FDB = 45°.

In △ACD:

In △BCD:

Thus, the total width of the river, AB = AC + BC = 144.34 + 250 = 394.34 ≈ 394 m.

Therefore, the width of the river is 394 m.

Question 15

The horizontal distance between two towers is 120 m. The angle of elevation of the top and angle of depression of the bottom of the first tower as observed from the top of the second tower is 30° and 24°, respectively. Find the height of the two towers. Give your answer correct to 3 significant figures.

Consider the diagram provided, where AB represents the first tower and CD stands for the second tower.

The horizontal distance between the two towers, EC or BD, is given as 120 m.

Let’s analyze △AEC:

The angle of elevation to the top of the first tower from the top of the second tower is 30°. Using the tangent of this angle:

This gives us:

Solving for AE, we have:

Substituting the value of EC:

Calculating further:

\Rightarrow AE = \dfrac{120}{1.732} = 69.28 \text{ m}.

Now, consider △EBC:

The angle of depression to the bottom of the first tower from the top of the second tower is 24°. Using the tangent of this angle:

This gives us:

Solving for EB, we have:

Substituting the value of EC:

\Rightarrow EB = 120 \times 0.445 = 53.4 \text{ m}.

From the diagram, we can see:

⇒ The height of the second tower, CD, is equal to EB, which is 53.4 meters.

⇒ The height of the first tower, AB, is the sum of AE and EB:

AB = AE + EB = 69.28 + 53.4 = 122.68 \approx 123 \text{ meters}.

Therefore, the heights of the two towers are 123 meters and 53.4 meters, respectively.

Question 16

The angles of depression of two ships A and B as observed from the top of a lighthouse 60 m high are 60° and 45° respectively. If the two ships are on the opposite sides of the lighthouse, find the distance between the two ships. Give your answer correct to nearest whole number.

Consider CD as the lighthouse.

The angles of depression to ships A and B are given as 60° and 45°, respectively. Since alternate interior angles are equal, we have:

∴ ∠DAC = ∠EDA = 60° and ∠DBC = ∠FDB = 45°.

Now, examine △ACD:

Next, analyze △BCD:

The total distance between the two ships, AB, is:

Hence, the distance between two ships = 95 m.