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At a point on level ground, the angle of elevation of a vertical tower is found such that it's tangent is 5/12 . On walking 19.2 mtrs towards the tower , the tangent of the angle of elevation is 3/4 . Find height of the tower?????

Here is the link for answer to your query!

https://www.meritnation.com/ask-answer/question/the-angle-of-elevation-a-of-a-vertical-tower-from-a-poin/some-applications-of-trigonometry/1556217

In the above link you just need to replace the 192 m by 19.2 m and rest of the conditions are same. 

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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

tanADB=AB/BD

tanADB = 3/4

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.

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