Right Triangles Class 9 ICSE — Chapter 24 Solutions
Chapter 24, Solution of Right Triangles, is where right triangles class 9 ICSE Maths comes together. The chapter defines what it means to solve a right-angled triangle, then gives you a ratio method for every case. You will find unknown sides, read unknown angles from tables, and split compound figures into right triangles. Section 24.1 carries the method and seven worked examples, collected with the rest of the Class 9 Maths book collection.
ICSE Class 9 Mathematics Chapter 24 Solution of Right Triangles PDF
What is inside Chapter 24
6 pages · 1 sections · 7 worked examples · 20 exercise questions
| Section | Title |
|---|---|
| 24.1 | INTRODUCTION |
- EXERCISE 24: 20 questions
24.1 INTRODUCTION
What solving a right triangle means
To solve a right-angled triangle means to find the remaining angles and the remaining sides. The triangle already has one right angle of 90°. Your job is the two other angles and the two other sides. The chapter opens with this definition, then lists the two cases in which a triangle can be solved. Every worked example that follows applies this same definition to a new figure.
A right triangle can be solved in exactly two situations.
- one side and one acute angle are given
- two sides of the triangle are given
Both cases work because the angle sum of a triangle is always 180°. Once one acute angle is known, the other is simply 90° minus it. Each case carries enough information to finish the whole triangle without guessing. You never need a third piece of data, and no other starting point exists. That is why the chapter groups every problem under these two cases.
The ratio method for an unknown side
This is the chapter’s core rule. Write the unknown side over the known side and set it equal to the corresponding trigonometrical ratio of the given angle. The ratio you choose depends entirely on which side is unknown.
If the unknown side is the perpendicular, use sin or tan. If it is the base, use cos or cot. The book stresses that cot means base over perpendicular. The most common mistake is swapping perpendicular and base. That turns tan into cot and changes the answer. You met these standard angle values in Chapter 23, Trigonometrical Ratios of Standard Angles.
Worked pattern. A right triangle has a 30° angle and a base of 15 cm. Find the perpendicular.
Step 1: The unknown side is the perpendicular and the base is known, so use tan 30°.
Step 2: Write \tan 30^\circ = \frac{\text{Perp.}}{\text{Base}}.
\tan 30^\circ = \frac{\text{Perp.}}{15} \Rightarrow \text{Perp.} = 15 \times \frac{1}{\sqrt{3}} = 5\sqrt{3}\ \text{cm}
Step 3: Use \sqrt{3} = 1.732.
\text{Perp.} = 5 \times 1.732 = 8.66\ \text{cm}
Final answer: The perpendicular is 5\sqrt{3}\ \text{cm}, about 8.66 cm.
Finding angles without calculating lengths
When two sides are given but no angle is, you can find the angle by ratio. Divide one given side by the other. Then match the result to a standard angle from the tables. You never compute a length first, and you never need the third side. This is the chapter’s second tool, after the ratio method.
Worked pattern. Suppose the perpendicular and the hypotenuse are in the ratio 1 to \sqrt{2}.
Step 1: Write the side ratio as a trig ratio. Here \frac{\text{Perp.}}{\text{Hypt.}} = \frac{1}{\sqrt{2}}.
Step 2: Match it to a standard angle.
\sin \theta = \frac{1}{\sqrt{2}} = \sin 45^\circ \Rightarrow \theta = 45^\circ
Final answer: The angle is 45^\circ.
The same idea works for tan. If the perpendicular equals the base, then \tan \theta = 1 and \theta = 45^\circ. The tables give you the angle directly. This is the quickest way to read an angle off the tables, and it appears throughout Exercise 24.
Compound figures split into right triangles
Several examples are not single right triangles. The method is to split the figure into right triangles and solve each one in order. The result of one triangle feeds the next. This works for a base split into BD and DC, a rhombus, a rocket path and a trapezium.
In a rhombus, the diagonals bisect each other at right angles. They also bisect the vertex angle. So the triangle at one corner carries a half angle and a right angle. The two half-diagonals are its legs, and the side of the rhombus is its hypotenuse.
The rocket rises 40 km vertically, then travels 40 km at 60° to the vertical. The angle between the second stage and the horizontal is 30°, because 90° − 60° = 30°. That step uses the complementary angle idea from Chapter 25.
In the trapezium, CD is parallel to BA. Co-interior angles give angle B = 180° − 120° = 60°. Then sin 60° finds the height CE, and cos 60° finds the extra base length BE. The area formula completes the problem.
A fresh worked example
Here is a complete example with fresh numbers. A ladder leans against a vertical wall. It makes an angle of 60° with the ground, and its top reaches 12 m up the wall. Find the length of the ladder.
Step 1: The wall is opposite the 60° angle, so it is the perpendicular. The ladder is the hypotenuse.
Step 2: Write the unknown over the known side and set it equal to sin 60°.
\sin 60^\circ = \frac{12}{L} \Rightarrow L = \frac{12}{\sin 60^\circ} = \frac{12}{\sqrt{3}/2} = \frac{24}{\sqrt{3}}
Step 3: Rationalise the denominator.
L = \frac{24}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{24\sqrt{3}}{3} = 8\sqrt{3}\ \text{m} \approx 13.86\ \text{m}
Final answer: The ladder is 8\sqrt{3}\ \text{m}, about 13.86 m long.
Rationalising the denominator removes the root from the bottom. It keeps the answer in the exact form 8\sqrt{3}\ \text{m}, which is what the exam expects. A decimal alone loses marks when the question wants a surd.
What to practise from this chapter
Exercise 24 holds 20 questions. They train four skills. Practise them in the order below.
- Ratio selection: questions 1 to 3 find a side or angle in a single right triangle. Pick the ratio first, then substitute.
- Finding a length: questions 4 and 5 find a side such as AD from given angles and a base.
- Two-triangle figures: questions 6 to 13 and 17 to 19 need an intermediate step. Solve one right triangle, then carry the result into the next.
- Word problems: questions 15 and 16 apply the method to a ladder and a kite on a 100 m string.
- The rhombus: question 20 gives a rhombus with perimeter 96 cm and an obtuse angle of 120°. The side is 24 cm. Halve the side and apply the 30-60-90 ratios to reach the diagonals.
Start with ratio selection, then attempt the compound figures, then finish with timed word problems. The question diagrams for Exercise 24 sit in the chapter PDF section of this page. The same question types run through the rest of the Class 9 book collection.
Frequently asked questions
What does solving a right angled triangle mean in this chapter?
Solving a right angled triangle means finding the remaining angles and the remaining sides. The triangle already has one right angle. You can solve it when one side and one acute angle are given, or when two sides are given. These two cases cover every start, as shown across the Class 9 course pages.
Which trig ratio should I use when one side and one acute angle are given?
Write the unknown side over the known side and set it equal to the corresponding ratio of the given angle. If the unknown side is the perpendicular, use sin or tan. If it is the base, use cos or cot. Cot is base over perpendicular.
How do I find an angle without calculating the other lengths?
Divide the two given sides and match the ratio to a standard angle. For example, \sin \theta = \frac{1}{\sqrt{2}} gives \theta = 45^\circ. You read the matching angle straight from the tables.
How do I handle a figure that is not a single right triangle?
Split the figure into right triangles and solve each in order. Use the result of the first triangle in the next. In a rhombus, the diagonals bisect each other at right angles and bisect the vertex angle.
What types of questions are in Exercise 24?
Twenty questions. They cover ratio selection, single-triangle sides, two-triangle figures, and word problems on ladders and kites. Question 20 revisits the rhombus. The chapter belongs to the wider ICSE library, and you can verify the syllabus on the official CISCE website.
Reference: ICSE Class 9 Mathematics textbook material, chapter 24: Solution of Right Triangles — CISCE-aligned.