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Question 3
The angle of elevation of the top of a tower from a certain point is 30. If the observer moves 20m towards the tower, the angle of elevation of the top increases by 15. Find the height of the tower.

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Solution

Let the height of the tower be h.

Also, SR = x m, PSR=θ

Given that, QS=20 m

And, PQR=30

Now, in ΔPSR,

tan θ=PRSR=hx

tan θ=hx

x=htan θ ...(i)

Now, in ΔPQR,

tan 30=PRQR=PRQS+SR

tan 30=h20+x

20+x=htan 30

20+x=h3

20+htan θ=h3(ii) [from Eq.(i)]

After moving 20 m towards the tower, the angle of elevation of the top increases by 15.

i.e. PSR=θ=PQR+15

θ=30+15=45

20+htan 45=h3 [from Eq(i)]

20+h1=h3

h(31)=20

h=2031.3+13+1 [by rationalization]

=20(3+1)31=20(3+1)2

=10(3+1)m

Hence, the required height of tower is 10(3+1)m.


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