Height and Distances - Solved Questions and Answers

Last Updated : 26 Jun, 2025

Question 1: From the top of a lighthouse, the angles of depression of two ships are 30 and 45 degrees. The two ships, as it was observed from the top of the lighthouse, were 100 m apart. Find the height of the lighthouse.

Solution:

Here, we can apply the formula Height = Distance/[cot(original angle) – cot(final angle)]
⇒ Height of the lighthouse = 100 / (cot 30 – cot 45)
100 / (√3 – 1) = 50 √3 + 50 m

Question 2: An 80 m long ladder is leaning on a wall. If the ladder makes an angle of 45 degrees with the ground, find the distance of the ladder from the wall.

Solution:

Here, cos θ = Base / Hypotenuse
⇒ cos 45 = Base / 80
⇒ Base = 80 cos 45 = 80 / √2= 40 √2
Thus, Distance of the ladder from the wall = 40 √2 m

Question 3: There are two poles, one on each side of the road. The higher pole is 54 m high. From the top of this pole, the angle of depression of the top and bottom of the shorter pole is 30 and 60 degrees respectively. Find the height of the shorter pole.

Solution:

Let AB and CD be the two poles.
Let AC = x m and CD = h mNow, in triangle ABC,
tan 60 = AB / AC
⇒ √3 = 54 / AC
⇒ AC = 18 √3 m
Clearly, AC = DE = 18 √3 m
In triangle BED,
tan 30 = BE / DE
⇒ BE = DE tan 30
⇒ BE = 18 √3 / √3 m
⇒ BE = 18 m
⇒ CD = AE = AB – BE
⇒ CD = 54 – 18 = 36 m
Therefore, the height of the shorter pole = 36 m

Question 4: From the top of a tower 100m high, a person observes that the angle of elevation of the top of another tower is 60° and the angle of depression of the bottom of the tower is 30°. Then the height (in meters) of the second tower is.

Solution:

Let AB be the first tower and CD be the second tower.
From figure,
AB=CE
Tan(30)= CE/BE
⇒ 1/√3=100/AC ---eq1
Tan(60)=DE/BE
BE=AC
⇒ √3=DE-100/AC----eq2
AC=DC-100/√3---eq3
putting eq3 in eq1
1/√3=100*√3/DC-100
DC-100=100√3*√3
DC=400m

Question 5: The angle of elevation of the top of a tower from point A on the ground is 45°. On moving 20 meters toward the tower, the angle of elevation of the top of the tower becomes 60°. Find the height of the tower (in meters).

Solution:

Let h be the height of the tower and x be the distance between the initial position and the foot of the tower.
From the right-angled triangle AOB, where O is the foot of the tower, we have:
tan(45°) = h / x ⇒ h = x
From the right-angled triangle COB, where C is the new position of the observer, we have:
tan(60°) = h / (x - 20) ⇒ h = (x - 20) × √3
Equating both expressions for h, we get:
x = (x - 20) × √3
Simplifying, we get:
x = 20 × (√3 + 1)
Therefore, the height of the tower is:
h = x = 20 × (√3 + 1) meters, which is approximately 45.72 meters (rounded to two decimal places).

Question 6: If the length of a pole is 12m and the angle of elevation from the top of the pole to point A on the ground is 45°, then find the distance of the pole from point A?

Solution:

Here AB = 12m
According to given trick, p, b, h is 1 : 1 : √2
Here p = 12m
And, p/b = BC/AC = 1/1
⇒ 12/AC = 1/1
So, AC = 12 m
So, the required value is 12m.
CASE – 3, When angle is 60° :-
We know that tan 60° = p/b = √3/1
So, h = √(3 + 1) = 2
Then, if an angle of elevation/depression is given as 60°, then the ratio of p, b, h is √3 : 1 : 2

Question 7: A pole is broken at some height and makes 60° to the ground and the distance between the bottom point of the pole and the point it touches the ground is 10m. What is the length of the pole?

Solution:

Let the pole is broken from point B and it touches point A on the ground.
Since angle of elevation is 60° so p, b, h is √3 : 1 : 2
Then, tan 60° = p/b = √3/1
⇒ BC/AC = √3/1
⇒ BC/10 = √3/1
⇒ BC = 10√3 m = 17.3
And, AB = √(BC2 + AC2) = √(300 + 100)m = 20m (Approx)
Since the pole is broken from the point B so the length of pole = BC + AB = (17.3 + 20)m = 37.3m = 37 m (approximately)
The required length is 37m.

Question 8: The shadow of a tree decreases by 15m when the sun’s altitude changes from 45° to 60°. Find the length of the tree.

Solution:

Let AB be the length of the tree and the length of shadow be x when the angle of elevation is 45°.
Then, the length of shadow when the angle of elevation is 60° = x – 15m
For angle 45° in triangle ABD, we have
AB/BD = 1/1 = AB/x
So, AB = x
Now, for angle 60° in triangle ABC, we have
AB/BC = √3/1 = x/(x – 15)
⇒ x = √3x – 15√3
⇒ (√3 – 1)x = 15√3
⇒ x = 15√3/(√3 – 1)m
The length of the tree is 15√3/(√3 – 1)m.

Question 9: A pole stands upright on the ground with the help of two wires to its top and affix to the ground on opposite sides. If the angle of elevation for both wires are 30° and 60° respectively and the length of the first wire is 8m, then find the length of another wire.

Solution:

Let AD be the length of the pole and AB and AC be the length of the first and second wires respectively.
In triangle ABD, we have
∠B = 30°
So, AB : AD = h : p = 2 : 1
⇒ 8/AD = 2/1
⇒ AD = 4m
Now, in triangle ADC, we have
∠C = 60°
So, AD : AC = √3/2
⇒ 4/AC = √3/2
⇒ AC = 8/√3 = 8√3/3 m
The required length of wire is 8√3/3 m.

Question 10: Angle of depression from a kite flying into the sky to a point on the ground is 45° and it changes to 60° when the kite flies 10m higher. Find the initial height of the kite.

Solution:

Let the kite was flying at a height of AB initially.
According to the question, in triangle ABC, we have ∠C = 45°
So, AB = BC = 1/1
Then, AB = BC
Now, in triangle ADC, we have ∠C = 60°
So, AD/BC = √3/1
⇒ (AB + 10)/AB = √3
⇒ AB + 10 = AB√3
⇒ AB(√3 – 1) = 10 m
⇒ AB = 10/(√3 – 1) = 5(√3 + 1) m
The required length is 5(√3 + 1)m.