The angle of elevation of the top of the tower as seen from a point on the ground is teetha such that tan theta = 3/4 from another point 240m away from this point on the line joining this point to the foot of the tower the angle of elevation of the top of the tower is such that tan theta =5/12 find the height of the tower
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Answered by
24
Let the height be h units
first distance from the foot of the tower to the point = x units
A/q
tanα = 3/4 = h/x
⇒3x=4h
also, tanβ =5/12 = h/(x +240)
⇒12h = 5x +1200
⇒3×3x = 5x +1200
⇒4x = 1200
⇒x =300 m
so height, h = (3×300)/4
⇒h =225 m
first distance from the foot of the tower to the point = x units
A/q
tanα = 3/4 = h/x
⇒3x=4h
also, tanβ =5/12 = h/(x +240)
⇒12h = 5x +1200
⇒3×3x = 5x +1200
⇒4x = 1200
⇒x =300 m
so height, h = (3×300)/4
⇒h =225 m
Answered by
7
Answer:
Let the height be h units
first distance from the foot of the tower to the point = x units
A/q
tanα = 3/4 = h/x
⇒3x=4h
also, tanβ =5/12 = h/(x +240)
⇒12h = 5x +1200
⇒3×3x = 5x +1200
⇒4x = 1200
⇒x =300 m
so height, h = (3×300)/4
⇒h =225 m
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