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: