Question

the angles of evelation of the top of a tower from two points at a distance of 4m and 9m from the base of the towr and in the same straight line with it are complementary. prove that the height of the tower is 6m.​

Answer 1

Given: 

BC = 4 cm, BD = 9cm.

$\sf{{\angle}ACB\:=\: (90^{\circ}\: -\: \theta)}$ and $\sf{\angle{ADB} \: = \: \theta}$

Prove: 

Height of tower i.e. AB = 6m

Proof:

In △ABC

=> tan (90° - Ø) = $\sf{\dfrac{AB}{BC}}$

=> cotØ = $\sf{\dfrac{AB}{BC}}$

=> $\sf{\dfrac{1}{tan\theta}}$ = $\sf{\dfrac{BC}{AB}}$

=> tanØ = $\sf{\dfrac{BC}{AB}}$ ____ (eq 1)

In △ABD

=> tanØ = $\sf{\dfrac{AB}{BD}}$ _____ (eq 2)

On comapring (eq 1) and (eq 2) we get

=> $\sf{\dfrac{BC}{AB}}$ = $\sf{\dfrac{AB}{BD}}$

Cross multiply them

=> BC × BD = AB × AB

=> 4 × 9 = AB²

=> 36 = AB²

=> AB = 6 m

•°• Height of tower is 6 m.

________________________________

Points to remember :-

tanØ = $\sf{\dfrac{P}{B}}$

cotØ = $\sf{\dfrac{B}{P}}$

Answer 2

$\bold{\underline{\underline </p><p>{Answer:}}}$

Given :

•The angle of elevation of a tower from two distinct points, 4m and 9m respectively from the base of the tower in the same straight line with it are complimentary.

•Height of the tower = 6 m

To prove :

•We have to prove that the height of the tower is 6 m.

Proof :

In Δ PQS,

Side QS = 4 m

$\angle{PSQ}$ = $\bold{(90-\theta)}$

$\rightarrow$ $\bold{tan\theta}$ = $\bold{\dfrac{Opposite\:side}{Adjacent\:side}}$

Opposite side,PQ = x m

[ This is also the height of the tower]

Adjacent side = QS = 4 m

$\rightarrow$ $\bold{tan\theta}$ = $\bold{\dfrac{x}{4}}$ ---> (1)

We know that :

• tan $\bold{(90-\theta)}$ = $\bold{cot\theta}$

•°•Substituting $\bold{cot\theta}$ for tan $\bold{(90-\theta)}$

$\rightarrow$ $\bold{cot\theta}$ = $\bold{\dfrac{x}{4}}$

$\bold{cot\theta}$ = $\bold{\dfrac{1}{tan\theta}}$

$\rightarrow$ $\bold{\dfrac{1}{tan\theta}}$ = $\bold{\dfrac{4}{x}}$

$\rightarrow$ $\bold{tan\theta}$ = $\bold{\dfrac{4}{x}}$ ---> (2)

Now, in Δ PRQ,

$\rightarrow$ $\bold{tan\theta\:={\dfrac{Opposite\:side}{Adjacent\:side}}}$

$\rightarrow$ $\bold{tan\theta\:={\dfrac{PQ}{QR}}}$

$\rightarrow$ $\bold{tan\theta\:={\dfrac{x}{9}}}$ ---> (3)

From equation 2 and equation 3,

$\rightarrow$$\bold{\dfrac{4}{x}}$ = $\bold{\dfrac{x}{9}}$

Cross multiplying,

$\rightarrow$ $\bold{4\times\:9\:=x\times\:x}$

$\rightarrow$ $\bold{36=x^2}$

$\rightarrow$ $\bold{\sqrt{36}=x}$

$\rightarrow$ $\bold{6=x}$

Initially we assumed x = PQ to be the height of the tower.

And we got the value of x = 6 m and that's what we were supposed to prove.

Hence proved.