A group of students of class X visited India Gate on an education

Q: A group of students of class X visited India Gate on an education trip. The teacher and students had interest in history as well. The teacher narrated that India Gate, official name Delhi Memorial, originally called All-India War Memorial, monumental sandstone arch in New Delhi, dedicated to the troops of British India who died in wars fought between 1914 and 1919. The teacher also said that India Gate, which is located at the eastern end of the Rajpath (formerly called the Kingsway), is about 138 feet (42 metres) in height.

A group of students of class X visited India Gate on an education trip. The teacher and students had interest in history as well.

1. What is the angle of elevation if they are standing at a distance of 42m away from the monument?

a) 30⁰b) 45⁰c) 60⁰d) 0⁰

2. They want to see the tower at an angle of 60^o. So, they want to know the distance where they should stand and hence find the distance.

a) 24.25 m. b) 20.12 m. c) 42 m. d) 24.64 m

3. If the altitude of the Sun is at 60^o, then the height of the vertical tower that will cast a shadow of length 20 m is

a) 20√3 m. b) $\frac{20}{\sqrt3}$ m. c) $\frac{15}{\sqrt3}$ m. d) 15√3 m

4. The ratio of the length of a rod and its shadow is 1:1. The angle of elevation of the Sun is

a) 30⁰b) 45⁰c) 60⁰d) 90⁰

5. The angle formed by the line of sight with the horizontal when the object viewed is below the horizontal level is

a) corresponding angle. b) angle of elevation.

c) angle of depression. d) complete angle

1. Angle of elevation at 42 m distance:

Height AB = 42 m

Distance from the base, BC = 42 m

Now, ∠ ACB is $\theta$ ,

∴ $\tan \theta = \frac {AB}{BC}$

By substituting values in above equation, we get:

$\tan \theta = \frac {42}{42} = 1$

∵ we know that tan 45⁰ = 1

∴ $\tan \theta = \tan 45$

∴ $\theta$ = 45⁰

Hence, option b) is correct.

2. Distance from base when angle is 60⁰:

From the adjoining figure, we can see that in ∆ ABC,

Given that the Height AB = 42 m

Also given that the Angle of elevation, ∠ACB = 60⁰

Let’s consider the distance from base is X m

Now, from the triangle ACB:

∴ $\tan ACB = \frac {AB}{BC}$

By substituting values in above equation, we get:

$\tan 60 = \frac {42}{\times} = 1$

∵ we know that $\tan 60 = \sqrt 3$

∴ $\sqrt 3 = \frac {42}{\times}$

∴ $\times = \frac {42}{\sqrt 3} = 14 \sqrt 3$ = 14 x 1.732 = 24.248

Hence, option a) is correct.

3. Height of the vertical tower:

From the adjoining figure, we can see that in ∆ ABC,

Given that the angle of elevation, ∠ACB = 60⁰

Also given that the length of shadow = 20 m

Let’s consider the height of the tower is X m

Now, from the ∆ ACB:

∴ $\tan ACB = \frac {AB}{BC}$

By substituting values in above equation, we get:

$\tan 60 = \frac {\times}{20}$

∵ we know that $\tan 60 = \sqrt 3$

∴ $\sqrt 3 = \frac {\times}{20}$

∴ X = $20 \sqrt 3$ m

Hence, option a) is correct.

4. Angle of elevation of the Sun:

we can see that in ∆ ABC,

Height AB = Distance from the base, BC

Now, let’s consider that the ∠ ACB is $\theta$ ,

∴ $\tan \theta = \frac {AB}{BC}$

By substituting values in above equation, we get:

$\tan \theta = \frac {AB}{AB} = 1$

∵ we know that tan 45⁰ = 1

∴ $\tan \theta = \tan 45$

∴ $\theta$ = 45⁰

Hence, option b) is correct.

5. Angle Name:

Since the angle between the horizontal line and the observation of the object from the horizontal line, when object is below the horizontal, is called angle of depression.