ICSE Class 9 Maths Chapter 24 Solution of Right Triangles

Solution Of Right Triangles

Simple 2-D Problems Involving One Right-angled Triangle

24.1 Introduction

To solve a right-angled triangle means, to find the values of remaining angles and remaining sides; when:

(i) one side and one acute angle are given.

(ii) two sides of the triangle are given.

Example 1

In a triangle ABC, right-angled at B, side BC = 20 cm and angle A = 30 degrees. Find the length of AB.

Solution

As shown alongside:

\(\frac{AB}{BC} = \cot 30°\)

Since Base/Perpendicular = cot

\(\frac{AB}{20} = \sqrt{3}\)

and, AB = 20 × \(\sqrt{3}\) = 20 × 1.732 cm = 34.64 cm

Answer: 34.64 cm

General Rule

In general, when one side and one acute angle of a right-angled triangle are given; we take:

\(\frac{\text{unknown side}}{\text{known side}} = \text{corresponding trigonometrical ratio of the given angle}\)

Example 2

Without calculating other lengths, use tables to find the angles x degrees and y degrees.

Solution

\(\sin x° = \frac{\text{Perp.}}{\text{Hyp.}} = \frac{5}{10} = \frac{1}{2} = \sin 30°\)

\(x° = 30°\)

Answer: 30 degrees

Similarly, \(\tan y° = \frac{\text{Perp.}}{\text{Base}} = \frac{10}{10} = 1 = \tan 45°\)

\(y° = 45°\)

Answer: 45 degrees

Teacher's Note

Understanding right triangles helps us calculate heights of buildings, distances to objects, and angles in navigation and surveying work.

Example 3

Use the information, given in the adjoining figure, to find:

(i) length of BD

(ii) angle C i.e. theta

(iii) length of BC.

Solution

(i) In right-angled triangle ABD,

\(\tan 45° = \frac{AD}{BD} \Rightarrow 1 = \frac{10}{BD} \Rightarrow BD = 10 \text{ cm}\)

Answer: 10 cm

(ii) In right-triangle ADC,

\(\sin \theta = \frac{AD}{AC} = \frac{10}{20} = \frac{1}{2} = \sin 30° \therefore \theta = 30°\)

Answer: 30 degrees

(iii) In right-triangle ADC,

\(\tan \theta = \frac{AD}{DC} \Rightarrow \tan 30° = \frac{10}{DC}\)

\(\Rightarrow \frac{1}{\sqrt{3}} = \frac{10}{DC} \Rightarrow DC = 10 \times 1.732 \text{ cm} = 17.32 \text{ cm}\)

\(BC = BD + DC = (10 + 17.32) \text{ cm} = 27.32 \text{ cm}\)

Answer: 27.32 cm

Teacher's Note

Surveyors use these multi-part triangle problems to measure land areas and determine distances between landmarks that cannot be directly measured.

Example 4

From the adjoining figure; find the length of AC.

Solution

Let DC = x m

\(\Rightarrow BC = BD + DC = (40 + x) \text{ m}\)

In triangle ABC, \(\tan 30° = \frac{AC}{BC}\)

\(\Rightarrow \frac{1}{\sqrt{3}} = \frac{AC}{40 + x} \Rightarrow 40 + x = AC \sqrt{3}\) - equation I

In triangle ADC, \(\tan 45° = \frac{AC}{DC} \Rightarrow 1 = \frac{AC}{x}\) and x = AC - equation II

Substituting x = AC in equation I, we get:

\(40 + AC = AC \sqrt{3} \Rightarrow AC \sqrt{3} - AC = 40\)

\(\Rightarrow AC (\sqrt{3} - 1) = 40\)

And, \(AC = \frac{40}{\sqrt{3} - 1} = \frac{40}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1}\)

\(= \frac{40(1.732 + 1)}{3 - 1} = 20 \times 2.732 \text{ m} = 54.64 \text{ m}\)

Answer: 54.64 m

Example 5

In a rhombus ABCD, length of each side is 10 cm and angle A = 60 degrees. Find the lengths of its diagonals AC and BD.

Solution

We know that the diagonals of a rhombus bisect each other at right angles and also bisect the angle of the vertex.

\(\Rightarrow OA = OC = \frac{1}{2} \text{AC}, OB = OD = \frac{1}{2} \text{BD}; \angle AOB = 90°\)

and \(\angle OAB = \frac{60°}{2} = 30°\). Also, given: AB = 10 cm.

In right triangle AOB:

\(\sin 30° = \frac{OB}{AB} \Rightarrow \frac{1}{2} = \frac{OB}{10} \Rightarrow OB = 5 \text{ cm}\)

\(\cos 30° = \frac{OA}{AB} \Rightarrow \frac{\sqrt{3}}{2} = \frac{OA}{10} \Rightarrow OA = 5\sqrt{3} \text{ cm} = 5 \times 1.732 = 8.66 \text{ cm}\)

Length of diagonal AC = 2 × OA = 2 × 8.66 = 17.32 cm

Answer: 17.32 cm

And, length of diagonal BD = 2 × OB = 2 × 5 cm = 10 cm

Answer: 10 cm

Teacher's Note

Rhombus properties are used in jewelry design and crystallography to understand geometric shapes with equal sides and symmetrical properties.

Example 6

In the given figure, a rocket is fired vertically upwards from its launching pad P. It first rises 40 km vertically upwards and then 40 km at 60 degrees to the vertical. PA represents the first stage of the journey and AB the second. C is a point vertically below B on the horizontal level as P, calculate:

(i) the height of the rocket when it is at point B.

(ii) the horizontal distance of point C from P.

Solution

Draw AD perpendicular to BC. Clearly angle BAD = 90 degrees - 60 degrees = 30 degrees

(i) In triangle ABD, \(\sin 30° = \frac{BD}{AB}\)

\(\Rightarrow \frac{1}{2} = \frac{BD}{40\text{ km}} \text{ i.e. } BD = 20 \text{ km}\)

The height of rocket when it is at point B = BC = BD + DC = 20 km + 40 km = 60 km

Answer: 60 km

(ii) In triangle ABD, \(\cos 30° = \frac{AD}{AB}\)

\(\Rightarrow \frac{\sqrt{3}}{2} = \frac{AD}{40 \text{ km}} \Rightarrow AD = 20\sqrt{3} \text{ km}\)

The horizontal distance of point C from P = PC = AD = 20 \sqrt{3} \text{ km} = 20 \times 1.732 \text{ km} = 34.64 \text{ km}\)

Answer: 34.64 km

Teacher's Note

Rocket trajectory calculations use trigonometry to determine optimal launch angles and distances for space missions and ballistic applications.

Example 7

In the given figure, AB perpendicular BC, DC perpendicular BC, BD perpendicular AC, angle D = 30 degrees and DC = 60 square root of 3 m. Find the length of AB.

Solution

In right triangle BCD,

\(\tan 30° = \frac{BC}{DC} \Rightarrow \frac{1}{\sqrt{3}} = \frac{BC}{60\sqrt{3}} \text{ i.e. } BC = 60 \text{ m}\)

In triangle BCD, angle BCD = 90 degrees

Also, angle DBC + 30 degrees + 90 degrees = 180 degrees

\(\Rightarrow\) angle DBC = 60 degrees

In triangle BOC, angle BOC + angle DBC + angle BCO = 180 degrees

90 degrees + 60 degrees + angle BCO = 180 degrees

and angle BOC = 30 degrees

In triangle ABC, \(\tan 30° = \frac{AB}{BC}\)

\(\Rightarrow \frac{1}{\sqrt{3}} = \frac{AB}{60}\)

\(\therefore AB = \frac{60}{\sqrt{3}} \text{ m} = \frac{60}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \text{ m}\)

\(= \frac{60\sqrt{3}}{3} \text{ m} = 20\sqrt{3} \text{ m}\)

Answer: 20 square root 3 m

Example 8

In the given figure, ABCD is a trapezium with angle C = 120 degrees, DC = 28 cm and BC = 40 cm. Find:

(i) AB

(ii) AD

(iii) the area of the trapezium.

Solution

Since, CD/BA and BC is transversal

\(\therefore\) angle B + angle C = 180 degrees - Co-interior angles

\(\Rightarrow\) angle B = 180 degrees - angle C = 180 degrees - 120 degrees = 60 degrees

Draw CE perpendicular to BA

In right triangle CBE,

\(\sin 60° = \frac{CE}{40 \text{ cm}}\)

\(\Rightarrow CE = 40 \times \frac{\sqrt{3}}{2} \text{ cm} = 20\sqrt{3} \text{ cm}\)

And, \(\cos 60° = \frac{BE}{40 \text{ cm}}\)

\(\Rightarrow BE = 40 \times \frac{1}{2} = 20 \text{ cm}\)

(i) AB = BE + AE = BE + CD = 20 cm + 28 cm = 48 cm

Answer: 48 cm

(ii) AD = CE = 20 \sqrt{3} \text{ cm}\)

Answer: 20 square root 3 cm

(iii) Area of trapezium = \(\frac{1}{2}\)(AB + CD) × AD = \(\frac{1}{2}\)(48 + 28) × 20\sqrt{3} cm² = 760\sqrt{3} cm²

Answer: 760 square root 3 square cm

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