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Question
The horizontal distance between two poles is 15 m. The angle of depression of the top of first pole as seen from the top of second pole is 30°. If the height of the second pole is 24 m, find the height of the first pole. Use`[sqrt3=1.732]`
Solution
Let AB and CD be two poles, where CD = 24 m.
It is given that angle of depression of the top of the pole AB as seen from the top of the pole CD is 30° and horizontal distance between the two poles is 15 m.
∴ ∠CAL = 30° and BD = 15 m.
To find: Height of pole AB
Let the height of pole AB be h m.
AL = BD = 15 m and AB = LD = h
Therefore, CL = CD − LD = 24 − h
Consider right ΔACL:
`tan angleCAL =\text{perpendicular}/\text{Base}=(CL)/(AL)`
`rArrtan 30^o=(24-h)/15`
`rArr 24-h=15/sqrt3`
`rArr24-h=5sqrt3`
`rArrh=24-5sqrt3`
`rArrh=24-5xx1.732` `[\text{Taking}sqrt3=1.732]`
`rArrh=15.34`
Therefore, height of the pole AB = h m = 15.34 m.
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