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# At a point on level ground, the angle of elevation of a vertical   tower is found to be such that its tangent is 5/12. On walking 192m towards the tower, the tangent of the angle of  .elevation is  3/4. .Find the height of the tower.

Asked by farhan.k , on 22/2/11

let height of the tower b h

tan theta = 5/12

h /192 + x = 5/12

12h = 5(192 + x)

12h = 960 + 5x

x = -960 + 12h/5  __ 1

tan phi = h/x

3/4 = h/x

x = 4h / 3._____2

1 = 2

-960+12h/5 = 4h/3

3(-960 + 12h) = 20h

-2880 + 36h = 20h

2880 = 16h

Therefore h = 2880 / 16 =  180 m

Posted by Devanshi Modha(DALMIA PUBLIC SCHOOL) on 22/2/11

Posted by farhan.kon 23/2/11

nswers

Let, AB is the height of the tower

here,CD = 19.2m

To find the length of AB

In tri. ABC,

tanABC = AB/BC

and tanABC = 5/12

So,

AB/BC - 5/12

AB/BD+CD = 5/12

AB/BD+19.2 = 5/12

12AB = 5(BD+19.2)

12AB/5 = BD+19.2

BD = 12AB/5 - 19.2

BD = 12AB - 96/5 ........(i)

In tri. ABD

So,

AB/BD = 3/4

5AB/12AB-96 = 3/4

4*5AB = 3(12AB-96)

20AB = 36AB-288

20AB - 36AB = -288

-16AB = -288

AB = -288/-16

AB = 14

Thus height of the tower is 14m.

Posted by richasinha62...(student), on 26/2/12

Posted by Yagna Agarwal(OAKRIDGE INTERNATIONAL SCHOOL) on 26/2/13