Question

The angle of elevation of the top of an unfinished vertical building on a ground from a point which is at a distance 100m from the base of the building is 45 degree.how much height of building much be raised show that it's angle of elevation from the point is 60 degree

Answer 1

Step-by-step explanation:

original height = tan45* width = 100m

tan60 = new height / dist

new height = 100√3

extra dist = 100√3-100 = 100(√3-1) = 71.21m approx

Answer 2

height the building must be raised  by 73 m so that it's angle of elevation from the same point be 60 degree ​

Step-by-step explanation:

let QR be the tower and S be the point of observation making an angle of elevation 45°

let PQ be the raised height so that the angle of elevation be 60°

in triangle  QRS

$tan 45 = \frac{QR}{RS}$

$1 = \frac{QR}{100}$

QR = 100 m

in triangle PRS

$tan 60 = \frac{PR}{RS}$

$\sqrt{3} = \frac{100+PQ}{100}$

$100\sqrt{3} = 100+PQ$

$PQ = 100(\sqrt{3} -1)$

putting the value of √3 = 1.73

100(0.73) = PQ

73 = PQ

PQ = 73 m

hence ,

height the building must be raised  by 73 m so that it's angle of elevation from the same point be 60 degree ​

#Learn more:

The angle of elevation of the top of an unfinished our the point distance from its base is 30 degree how much higher much the tower b raised so that it angle of elevation at the same point is 60 degree

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