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.​

Answer 1

Answer:

the answer is solved in pic

Answer 2

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,

$\large \sf \frac{AQ}{QR}$ = tanθ

$\large \sf \frac{AQ}{4}$ = tanθ

2nd Equation :-

In ∆AQS,

$\large \sf \frac{AQ}{SQ}$ = tan(90 - tanθ)

$\large \sf \frac{AQ}{9}$ = cotθ

Next...

On multiplying equations (1st) and (2nd), we obtain;

$\large \sf (\frac{AQ}{4})(\frac{AQ}{9})$ = (tanθ) · (cotθ)

$\large \sf \frac{AQ²}{36}$ = 1

AQ² = 36

AQ = √36 ± 6

However, height cannot be negative.

Therefore, the height of the tower is 6 m.