Question

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Answer 1

Q. The angle of elevation of the top of a tower from two points at a distance of 4m & 9m from the base of the tower & in the same straight line with it are complementary. Prove that the height of the tower is 6m.

See the attachment

Answer:

tan a=h/4 _______i

tan b=h/9

Given that a+b=90

b=90-a

tan b =h/9

tan 90-a=h/9

By complementary angles

tan 90-a = cot a

So, cot a=h/9

cot a = 1/tan a

1/tan a=h/9

tan a=9/h _______ii

9/h=h/4

h²=36

h=6

Answer 2

Answer:

Let AB be the tower.

Let P and Q be the points at distance of 4m and 9m respectively.

From fig, PB = 4m, QB = 9m.

Let angle of elevation from P be α, and angle of elevation from Q be β.

Given that α and β are complementary. Thus, α + β = 90

(Two angles are complementary if their sum equals 90°.)

In triangle ∆ABP,

tan α = ABBP

………. (1)

In triangle ABQ,

tan β = ABBQ

Now, tan (90 – α) = tan β = ABBQ

(Since, α + β = 90)

⟹cot α = ABBQ

(tan (90-θ) = cot θ) and tan θ = 1cot θ

⟹1tan α

= ABBQ

So, tan α = BQAB

…….. (2)

From (1) and (2)

ABBP

= BQAB

⟹AB2

= BQ x BP

We know that, PB = 4m, QB = 9m.

⟹AB2

= 4 x 9

⟹AB2

= 36

Therefore, AB = 6.

Hence, the height of the tower is 6m.

Note: In order to solve this type of question the key is to draw an appropriate figure and establish the relation between what we want with what we have. Given angles are complementary is the biggest hint to break down the problem. Then we solve the relations for an answer.