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ICSE Class 10 Answered

the horizontal distance between two towers is 150 m the angle of depression of the top of one tower as observed from the top of other tower which is 120m in height is 30 degree find the height of the first tower
Asked by 410906486004709 | 05 Feb, 2018, 06:48: PM
answered-by-expert Expert Answer
begin mathsize 16px style Let space AB space and space CD space be space the space two space towers.
Height space of space tower space AB equals 120 space straight m
Horizontal space distance space between space two space towers equals BD equals 150 space straight m
In space triangle space ACE comma
tan space 30 equals AE over EC
rightwards double arrow fraction numerator 1 over denominator square root of 3 end fraction equals AE over 150
rightwards double arrow AE equals fraction numerator 150 over denominator square root of 3 end fraction equals fraction numerator 50 cross times 3 over denominator square root of 3 end fraction equals 50 square root of 3 space straight m
Now comma space EB equals AB minus AE equals 120 minus 50 square root of 3 equals CD
rightwards double arrow CD plus 120 minus 50 square root of 3 space straight m
end style
Answered by Rashmi Khot | 06 Feb, 2018, 10:06: AM

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