Height and Distance - Solved Questions and Answers

Height and Distance is a branch of trigonometry that calculates vertical heights, horizontal distances, and angles of elevation or depression using right-angled triangles and trigonometric ratios.

• Angle of Elevation: Upward angle from horizontal to an object above.
• Angle of Depression: Downward angle from horizontal to an object below.

Formulas:

Height = Distance × tanθ

Distance = Height / tanθ

Height and Distance questions and answers are provided below for you to learn and practice.

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(60°) = h / x ⇒ h = x

From the right-angled triangle COB, where C is the new position of the observer, we have:
tan(45°) = h / (x - 20) ⇒ h = (x - 20) × √3

Equating both expressions for h, we get:
√3x = x + 20
√3x - x = 20
x(√3-1) = 20
x = 20/(√3-1)

After rationalization:
x = 20 × /(√3 - 1) × [(√3 + 1)/(√3 - 1)]

Therefore, the height of the tower is:
h = x = 10 × (√3 + 1) meters, which is approximately 47.32 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:

Pole height (opposite side) = 12 m

Angle of elevation = 45°

tan45° ​= 12​/CA

1 = 12/CA

CA= 12m
Therefore, the distance of the pole from point A is 12m

Question 7: 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 8: 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 9: 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.

Question 10: If a pole 6 m high casts a shadow 2√3 m long on the ground, find the Sun’s elevation. 

Solution: 

Let AB be the pole which is of height 6 m and BC be the shadow of the building 2√3.

Now, in ∆ ABC,

tan θ = AB / BC
⇒ tan θ = 6 / 2√3

Now, simplifying using rationalization

tan θ = (3 / √3) × (√3 / √3)
⇒ tan θ = 1 / √3
⇒ θ = tan^-1(1 / √3)

Hence, θ = 60^o

Therefore, sun's elevation from the ground is 60^o.

Question 11: An observer 1.5 m tall is 20.5 m away from a tower 22 m high. Determine the angle of elevation of the top of the tower from the eye of the observer.

Solution: 

Let PQ be the height of the observer of 1.5 m.

Let AB be the height of the tower of 22 m.

And, let QB be the horizontal distance between the observer and the tower  

PQ = MB = 1.5 m
Thus, AM = AB - MB
⇒ AM = 22 - 1.5 = 20.5

Now, in ∆APM,

tan θ = AM / PM
⇒ tan θ = 20.5 / 20.5
⇒ tan θ = 1
⇒ θ = tan^-1(1 )

Hence, θ = 45^o

Therefore, the angle of elevation of the top of the tower from the eye of the observer is 45^o

Question 12: An airplane is flying h meters above the ground. At a particular instant, the angle of elevation of the plane from the eyes of a boy sitting on the ground is 60°. After some time, the angle of elevation changed to 30°. Find the distance covered by the plane during that time assuming it travelled in a straight line.

Solution:

Let x be the horizontal distance between the observer and plane at the first instant.

Let y be the horizontal distance between the observer and plane at the second instant.

And, BA = CD = h, Now in ∆OAB,

tan 60° = AB / OA
√3 = h / x
x = h / √3

In ∆ OCD,

tan 30° = CD / OD
1/√3 = h / (x+y)
x + y = √3h

Distance travelled by plane = AD = y

(x + y) − x = √3h − h / √3
y = (2 / √3)h

So, if the airplane is flying h meters above the ground, it would travel for (2/√3) h meters as the angle of elevation changes from 60° to 30°.

Question 13: From the top of the tower 30 m height a man is observing the base of a tree at an angle of depression measuring 30 degrees. Find the distance between the tree and the tower.

Solution: 

In the above diagram AB represents the height of the tower, BC represents the distance between the foot of the tower and the foot of the tree.

Now we need to find the distance between the foot of the tower and the foot of the tree (BC). For that as angle of depression is given so by vertically opposite angle property of triangle ∠CAD = ∠BCA

In ∆BCA,

tan θ = Opposite side / Adjacent side
⇒ tan 30° = AB / BC
⇒ 1/√3 = 30 / BC
⇒ BC = 30√3
⇒ BC = 30 (1.732)      [Approximately]
⇒ BC = 51.96 m

So, the distance between the tree and the tower is 51.96 m.

Question 14: From the top of a building 30 m high, the top and bottom of a tower are observed to have angles of depression 30° and 45° respectively. Find the height of the tower.

Solution:

Let AB be the building and CD be the tower.

The angle of depressions is given 30° and 45° to the top and bottom of the tower. So by vertically opposite triangle property ∠FBD = ∠EDB and ∠FBC = ∠ACB.

Now, Calculate the horizontal distance x using the angle of depression to the bottom of the tower 45°:

x = 30/tan45 = 30 = 30 meters

Calculate the height of the tower h using the angle of depression to the top of the tower (30°):

tan 30 = 30 - h/30

h = 30 - 30 × tan30

h = 30 - 30 × 1/√3 = 30 - 17.32 = 12.68 meters

So, the height of the tower is approximately 12.68 meters.

Also Check:

• Trigonometric Ratios
• Application of Trigonometry in Real Life
• Trigonometric Equations