# How To Solve Height And Distance Questions Quickly

How to solve Height and Distance Questions Quickly

In geometry Heights and Distances are the fundamental Concept that play a crucial role in real world application. This topic is of high importance in placement exams as well. In this page we are going to discuss about how we can easily and quickly solve heights and distance problems.

### How to Solve Height and Distance  Question Quickly

• Angle of elevation: The angle of elevation of an object as seen by an observer is the angle between the horizontal and the line from the object to the observer’s eye
• Angle of Depression: If the object is below the level of the observer, then the angle between the horizontal and the observer’s line of sight is called the angle of depression.

### Type 1: Find the distance/height/base/length when angle is given

Question 1. Radhika is 2 m tall and is standing 10$\sqrt{3}$ m away from a tower. The angle of elevation from her eye to the top of the tower is 30º. Find the height of the tower?

Options.

A. 15 m

B. 16 m

C. 14 m

D. 12 m

Solution :

DC = AB = 2m

AD = BC =10 $\sqrt{3}$

tan 30°=$\frac{ED}{AD}$

$\frac{1}{\sqrt{3}}$ = $\frac{ED}{10 \sqrt{3}}$

ED =10

EC = ED + DC = 10+2 = 12m.

Correct option: D

Question 2. The angle of elevation of a ladder inclined against a wall is 60°. The base of the ladder is at distance of 5.2 m away from the wall. Find the length of the ladder?

Options.

A. 12.3 m

B. 11.4 m

C. 10.4 m

D. 9.2 m

Solution:

QRP = 60°

QR = 4.6

cos 60° = $\frac{QR}{PR}$

$\frac{1}{2}$= $\frac{QR}{PR}$

PR = 2 × QR

PR = 2 × 5.2

PR = 10.4 m

Correct option: C

Question 3. A ship was floating on water. The deck of the ship was 10m above water level. Yug was standing on the deck of a ship. He observes the angle of elevation of the top of a light house as 60° and the angle of depression to the foot of lighthouse as 30°. Find the height of the light house?

Options.

A. 30m

B. 40m

C. 35 m

D. 20m

Solution:

In ΔSQR,

tan 30° = $\frac{10}{QR}$

$\frac{1}{\sqrt{3}}$ = $\frac{10}{QR}$

QR = 10 $\sqrt{3}$ m

In ΔPQR

tan 60° = $\frac{PR}{QR}$

$\sqrt{3}$ = PR/10√3

PR = 30m

h = 30m

Therefore, the height of the light house = PS = RS + PR

= 10 + 30 = 40m

Correct option: B

### Type 2: Find the angle when distance/height/base/length is given

Question 1. Find the angle of elevation of the sun, when the length of the shadow of an apple tree is equal to the height of an apple tree?

Options.

A. 45°

B. 60°

C. 30°

D. 90°

Solution:

Given in the question AB = BC

Let ABC = θ

tan θ = $\frac{AB}{AC}$ = 1 (since AB= BC)

tan θ = 1

θ = 45°

Correct option: A

Question 2. Find the angle of elevation of the sun, when the length of the shadow of a tree is √3 times the height of the tree?

Options.

A. 45°

B. 60°

C. 30°

D. 90°

Solution:

Let the height of tree AB be h

Since we are given that The shadow of tree is $\sqrt{3}$ times its height

Length of shadow = $\sqrt{3}$h

To Find angle of elevation of sun we will use trigonometric ratio

In ΔABC

Hence the angel of elevation is 30 °

Correct option: C

Question 3. From the figure given below, find the value of θ?

Options.

A. 45°

B. 30°

C. 90°

D. 60°

Solution:

In the given diagram,

Perpendicular (PQ) = 1

Base (QR) = $\sqrt{3}$

Therefore, tan θ = $\frac{1}{\sqrt{3}}$

θ = 30°

Correct option: B

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