Q: The angle of elevation of the sun when the shadow of a pole $h$ meters high is $\sqrt{3}h$ meters long is:

Let the angle of elevation of the sun be $\theta$. The height of the pole (opposite side) is $h$ and the length of the shadow (adjacent side) is $\sqrt{3}h$. Therefore, $\tan \theta = \frac{\text{Height}}{\text{Length of shadow}} = \frac{h}{\sqrt{3}h} = \frac{1}{\sqrt{3}}$. Since $\tan 30^\circ = \f

Question 1

The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. The height of the tower is:

Accepted Answer

Let the height of the tower be $h$ m. The distance from the foot of the tower is $30$ m. In the right-angled triangle, $	an 30^\circ = \frac{	ext{Height}}{	ext{Base}} = \frac{h}{30}$. Since $	an 30^\circ = \frac{1}{\sqrt{3}}$, we get $\frac{1}{\sqrt{3}} = \frac{h}{30}$, which implies $h = \frac{

Question 2

If the height of a tower and the distance of the point of observation from its foot are both increased by 10%, then the angle of elevation of its top:

Accepted Answer

Let the original height be $h$ and the original distance be $x$. The angle of elevation $	heta$ is given by $	an 	heta = \frac{h}{x}$. When both height and distance are increased by 10%, the new height is $1.1h$ and the new distance is $1.1x$. The new angle of elevation $	heta'$ is given by $	a

Question 3

The angle of elevation of the sun when the shadow of a pole $h$ meters high is $\sqrt{3}h$ meters long is:

Accepted Answer

Let the angle of elevation of the sun be $	heta$. The height of the pole (opposite side) is $h$ and the length of the shadow (adjacent side) is $\sqrt{3}h$. Therefore, $	an 	heta = \frac{	ext{Height}}{	ext{Length of shadow}} = \frac{h}{\sqrt{3}h} = \frac{1}{\sqrt{3}}$. Since $	an 30^\circ = \f

Question 4

A ladder 15 m long just reaches the top of a vertical wall. If the ladder makes an angle of 60° with the wall, then the height of the wall is:

Accepted Answer

Let the height of the wall be $h$. The length of the ladder is the hypotenuse, which is $15$ m. The angle made by the ladder with the vertical wall is $60^\circ$. Therefore, $\cos 60^\circ = \frac{	ext{Height of the wall}}{	ext{Length of the ladder}} = \frac{h}{15}$. Since $\cos 60^\circ = \frac{1

Question 5

The angle of depression of a car parked on the ground from the top of a 75 m high tower is 30°. The distance of the car from the base of the tower is:

Accepted Answer

The height of the tower is $75$ m. The angle of depression is equal to the angle of elevation, which is $30^\circ$. Let the distance of the car from the base of the tower be $x$. Then, $	an 30^\circ = \frac{	ext{Height}}{	ext{Distance}} = \frac{75}{x}$. Since $	an 30^\circ = \frac{1}{\sqrt{3}}$,

Applications of Trigonometry