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 .

get 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 . from screen.

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 be

Question 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 ​

Was this answer helpful?

144 11

स्रोत : www.toppr.com

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 8NCERT 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

स्रोत : www.cuemath.com

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

RELATED VIDEOS

Classification of Triangles

MATHEMATICS Watch in App EXPLORE MORE

Classification of Triangles Based on Angles

Standard VII Mathematics

स्रोत : byjus.com