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Question

Question 7
The shadow of a tower standing on a level plane is found to be 50 m longer when Sun’s elevation is 30 than when it is 60. Find the height of the tower.

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Solution

Let the height of the tower be h and RQ = x m.

Given that, PR = 50 m

And, SPQ=30, SRQ=60

Now, in ΔSRQ, tan 60=SQRQ

3=hxx=h3 ...(i)

And, in ΔSPQ,tan 30=SQPQ=SQPR+RQ=h50+x



13=h50+x

3.h=50+x

3.h=50+h3 [from Eq. (i)]

(313)h=50

(31)3h=50

h=5032h

Hence, the required height of the tower is 253 m.

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