# the angle of elevation of the top of a tower from two points distant s and t from its foot are complementary. prove that the height of the tower is st .

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## The angle of elevation of the top of a tower from two points distant s and t from its foot are complementary. Prove that the height of the tower is √(s t) .

Click here👆to get an answer to your question ✍️ The angle of elevation of the top of a tower from two points distant s and t from its foot are complementary. Prove that the height of the tower is √(s t) .

The angle of elevation of the top of a tower from two points distant s and t from its foot are complementary. Prove that the height of the tower isLet complementary angles beQuestion st . Medium Open in App Solution Verified by Toppr α and 90−α tanα= s h .....(i) tan(90−α)=cotα= t h ....(ii)

multiply (i) and (ii)

tanα.cotα= s h . t h ts h 2 =1 h= st

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## The angle of elevation of the top of a tower from two points distant s and t from its foot are complementary. Prove that the height of the tower is √st

The angle of elevation of the top of a tower from two points distant s and t from its foot are complementary. Prove that the height of the tower is √st - The angle of elevation of the top of a tower from two points distant s and t from its foot are complementary. It is proven that the height of the tower is √st

## The angle of elevation of the top of a tower from two points distant s and t from its foot are complementary. Prove that the height of the tower is √st

**Solution:**

Given, the angle of elevation of the top of a tower from two points distant s and t from its foot are complementary.

We have to prove that the height of the tower is √st.

Let AC be the height of the tower

AC = h units

P and B are the points of observation.

Given, PC = t units BC = s units

Given, angle of elevation is complementary

∠ABC = θ ∠APC = 90° - θ In triangle ABC, tan θ = AC/BC

tan θ = h/s ------------------ (1)

In triangle APC,

tan (90°- θ) = AC/PC

By using trigonometric ratio of complementary angles,

tan (90° - A) = cot A

So, tan (90°- θ) = cotθ

cot θ = AC/PC

cot θ = h/t -------------------- (2)

Multiplying (1) and (2),

tan θ × cot θ = (h/s)(h/t)

We know that tan A × cot A = 1

So, 1 = h²/st h² = st Taking square root, h = √st

Therefore, the height of the tower is √st.

**✦ Try This:**The angles of elevation of the top of a rock at the top and foot of a 100 m high tower, at respectively 30° and 45°. Find the height of the rock.

**☛ Also Check:**NCERT Solutions for Class 10 Maths Chapter 8

**NCERT Exemplar Class 10 Maths Exercise 8.4 Problem 6**

## The angle of elevation of the top of a tower from two points distant s and t from its foot are complementary. Prove that the height of the tower is √st

**Summary:**

The angle of elevation of the top of a tower from two points distant s and t from its foot are complementary. It is proven that the height of the tower is √st

**☛ Related Questions:**

The shadow of a tower standing on a level plane is found to be 50 m longer when Sun’s elevation is 3 . . . .

A vertical tower stands on a horizontal plane and is surmounted by a vertical flag staff of height h . . . .

If tanθ + secθ = l, then prove that secθ = l2+1/2l

## The angle of elevation of the top of a tower from two points distant s and t from its foot are complementary. Prove that the height of the tower is √st.

The angle of elevation of the top of a tower from two points distant s and t from its foot are complementary. Prove that the height of the tower is √st.

Byju's Answer Standard VII Mathematics

Classification of Triangles Based on Angles

The angle of ... Question

The angle of elevation of the top of a tower from two points distant

s and t

from its foot are complementary. Prove that the height of the tower is

st . Open in App Solution

**Step 1: Draw the diagram**

Let BC=s ; PC=t

be the distance of given points.

Let the height of the tower be

AC=h

∠ABC=θ and ∠APC=90°-θ

(∵ the angle of elevation of the top of the tower from two points P and B are complementary)

**Step 2: Proof**

In △ABC ⇒tanθ=ACBC

⇒tanθ=hs---------(i)

Now, In △ACP ⇒tan90°-θ=ACPC

=cotθ=ht--------(ii)

By multiplying the equations (i) and (ii) we obtain

⇒tanθ×cotθ=hs×ht

⇒ 1=h2st ∵tanθ=1cotθ

⇒ st=h2 ⇒ h=st

Hence, the height of the tower is

st Suggest Corrections 1

SIMILAR QUESTIONS

**Q.**The angle of elevation of top of a tower from two points distant

s and t

from its foot are complementary. Prove that the Height of the tower is

√ s t

**Q.**The angle of elevation of the top of a tower from two points distant

a and b

from the base and in the same straight line with it are complementary. The height of the tower is

**Q.**The angle of elevation of the top of a tower as seen from two points

A & B

situated the same line and at distance '

p ' and ' q

' respectively from the foot of the tower are complementary, then height of the tower is

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