WBBSE Class 10 Maths Trigonometry Chapter 4 Application Of Trigonometric Ratios Height And Distance Multiple Choice Questions

WBBSE Class 10 Maths Trigonometry Chapter 4 Application Of Trigonometric Ratios Height And Distance Multiple Choice Questions

Example 1. If the angle of elevation of the top of the mobile tower from a distance of 10 metres from its food is 60°, then the height of the tower is 

1. 10 m
2. 10√3 m
3. \(\frac{10}{\sqrt{3}}\) m
4. 100 m

Solution:

⇒ Let the height of tower (AB) = x m. and angle of elevation of the top of mobile tower is ∠ACB where ∠ACB = 60°

⇒ BC = 10 m

⇒ From ΔABC, ∠ABC = 90°

Read And Learn Also WBBSE Class 10 Maths Multiple Choice Questions

∴ \(\frac{AB}{BC}\) = tan 60°

\(\frac{x}{10}\) = √3

⇒ x = 10√3

∴ The height of the tower is 10√3 m

∴ The correct answer is 2. 10√3 m

Example 2. A value of θ is

1. 30°
2. 45°
3. 60°
4. 75°

Solution: ΔABC, ∠ABC = 90°

∴ \(\tan \theta=\frac{A B}{B C}=\frac{5}{5 \sqrt{3}}\)

⇒ \(\tan \theta=\frac{1}{\sqrt{3}}=\tan 30^{\circ}\)

∴ θ = 30°

∴ The correct answer is 1. 30°

Class 10 Maths Trigonometry Chapter 4 MCQs

Example 3. At what angle an observer observes a box lying on ground from the roof of three storied building, so that the height of the building is equal to the distance of the box from the building, then the angle is

1. 15°
2. 30°
3. 45°
4. 60°

Solution: Let height of building is AB and the distance of the box from the building is BC

∴ AB = BC

⇒ Let required angle is ∠ACB

⇒ From ΔABC, ∠ABC = 90°

⇒ tan ∠ACB = \(\frac{AB}{BC}\)

⇒ tan ∠ACB = \(\frac{AB}{AB}\) [as AB = BC]

⇒ tan ∠ACB = 1 = tan 45°, ∠ACB = 45°

∴ The correct answer is 3. 45°

Example 4. Height of tower is 100√3 metre. The angle of elevation of the top of a tower from a point at a distance of 100 metres of foot of tower is

1. 30°
2. 45°
3. 60°
4. None of these

Solution: Let height of tower is AB the angle of elevation of the top of a tower from a point C at a distance BC is θ

∴ AB = 100√3 metre and BC = 100 metre

⇒ In ΔABC, ∠ABC = 90°

∴ tanθ = \(\frac{A B}{B C}=\frac{100 \sqrt{3}}{100}=\sqrt{3}\)

⇒ tanθ = tan 60° ⇒ θ = 60°

∴ The correct answer is 3. 60°

Trigonometric Equations Mcqs Class 10

Example 5. If the length of shadow on the ground of a post is √3 times of its height, the angle of elevation of the sun is

1. 30°
2. 45°
3. 60°
4. none of these

Solution: Let, the length of post is AB and the length of shadow on the ground of the post is BC

∴ BC = √3 AB

Let the angle of elevation of the sun is θ

⇒ In ΔABC, ∠ABC = 90°

⇒ tanθ = \(\frac{A B}{B C}=\frac{A B}{\sqrt{3} A B}=\frac{1}{\sqrt{3}}\)

⇒ tanθ = tan 30°

⇒ θ = 30°

∴ The correct answer is 1. 30°

Example 6. If the angle of elevation of a light post of height 20 m is 45°, then the length of shadow of the post is

1. 20 m
2. 20√3 m
3. \(\frac{20}{\sqrt{3}} \mathrm{~m}\)
4. 10 m

Solution: The height of light post (AB) is 20 m and length of shadow of the post is BC;

∠ABC = 90°, ∠ACB = 45°

⇒ In ΔABC, \(\frac{AB}{BC}\) = tan 45°

⇒ \(\frac{20 M}{BC}\) = 1

⇒ BC = 20 m.

∴ The correct answer is 1. 20 m

Class 10 Maths Chapter 4 Trigonometry Solutions

Example 7. ∠ABD = 90°, AB = 10 m, the angle of elevation of point A at D and C are 45° and 30°, if DC = x m, then the value of x is

1. 10(√2 +1)
2. 10(√3+1)
3. 10(√3-1)
4. 10(√2 -1)

Solution: In ΔABD, \(\frac{AB}{BD}\) = tan 45°

⇒ \(\frac{10 M}{BD}\)

⇒ BD = 10 m

⇒ In ABC, \(\frac{AB}{BC}\) = tan 30°

⇒ \(\frac{10}{10+x}=\frac{1}{\sqrt{3}}\)

⇒ 10 + x= 10√3

⇒ x = 10(√3-1)

∴ The correct answer is 3.

WBBSE Class 10 Maths Trigonometry Notes

Example 8. If the ratio of the length of shadow of the tower and the height of the tower is 1: 1 then the angle of elevation of the sun is

1. 30°
2. 45°
3. 60°
4. 90°

Solution: AB is the height and BC is length of the shadow of tower AB.

⇒ AB : BC = 1 : 1

⇒ \(\frac{AB}{BC}\) = 1 ⇒ AB = BC

⇒ In ΔABC, ∠ABC = 90°

⇒ tan ∠ACB = \(\frac{AB}{BC}\) = \(\frac{AB}{AB}\) = 1

⇒ tan ∠ACB = tan 45°

⇒ ∠ACB = 45°

The angle of elevation is 45°

∴ The correct answer is 2. 45°