Q1. Help...!!
The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m. from the base of
the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.
Answers
Explanation:
Question :-
The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m. from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.
Solution :-
Let AQ be the tower and R, S are the points 4m, 9m away from the base of the tower respectively.
The angles are complementary. Therefore, if one angle is θ, the other will be 90 - θ.
1st Equation :-
In ΔAQR,
2nd Equation :-
In ∆AQS,
Next...
On multiplying equations (1st) and (2nd), we obtain;
AQ² = 36
AQ = √36 ± 6
However, height cannot be negative.
Therefore, the height of the tower is 6 m.
Thank You*
Height of Tower = 6 metre
Given Terms:
- Angles of elevation at two points are complementary.
- Distance of two points from the base of tower is 4 and 9 m.
Need To Prove:
- Height of tower is 6 metre.
Proof: Let the height of tower be x metre. In the attached diagram we have
- AB is tower of 6 m.
- BC is first point = 4 m.
- BD is second point = 9 m.
- ∠ACB = θ
- ∠ADB = 90° – θ
We'll use three formulae here
- Tanθ = Perpendicular/ Base
- Tan(90 – θ) = cotθ
- tanθ × cotθ = 1
In triangle ACB
➥ tanθ = AB/BC
➥ tanθ = h/4
➥ 4tanθ = h equation (i)
In triangle ADB
➥ tan(90 – θ) = AB/BD
➥ cotθ = h/9
➥ 9cotθ = h equation (ii)
Now multiply both the equations , we will get
➼ 4tanθ × 9cotθ = h × h
➼ 36 = h²
➼ √36 = h
➼ 6 = h
Here it is proved that the height of tower is 6 metre.