A TV tower stands vertically on a bank of a canal, with a height of10✓13 m. From a point on the
other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From
another point on the line joining this point to the foot of the tower, the angle of elevation of the top
of the tower is 30°. Find the distance between the
opposite bank of the canal and the point with 30°
angle of elevation.
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Answer:
let BC= y metre ,DB= x metre
hieght of tower =10√3m
Step-by-step explanation:
In ∆ABD,
AB/DB=tan30°
10√3/x=1/√3
x =10√3×√3
x=10×3,
x=30
now, In ∆ABC
AB/BC=tan60°
10√3/y=√3
10√3=y√3
10√3/√3=y
now we have to find DC
therefore, DB-CB =DC
30-10√3/√3=DC
DC=20m ans.
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