Puneet and Karim, both are standing in front of their school building. The angle of elevation of the top of the building for Puneet is 45° and that for Karim is 60°. Find the distance between Puneet and Karim, if the height of the school building is 30 m. (Take value of √3 = 1.732 and assume that the heights of Puneet and Karim are negligible.) pls give the complete answer of this very question....dont give links

Dear student
In ABCtan45°=ABBC=30y=1y=30m Now in ADBtan60°=ABDB=30x=3x=303=301.732=17.32So the distance between puneet and karim =y-x=30-17.32=12.68m

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Distance of Puneet from school = (30/tan 45o) m = 30/1 m = 30 m = x (let)
Distance of Karim from school = (30/tan60o) m = 30/(1.732) m = 17.32 m approx. = y (let)
Thus, distance between Puneet and Karim = x-y = (30 - 17.32) m = 12.68 m (Answer)
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