From a point P on the ground the angle of elevation of the top of a tower is
30° and that of the top of a flag staff fixed on the top of the tower, is 60°. If
the length of the flag staff is 5 m, find the height of the tower. Also find the
distance of point P from the foot of the tower. (Take √3 =1.7)
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Let , the height of tower (DC) be " x "
In Δ ABC
\tt \implies \tan(60) = \frac{5 + x}{bc}⟹tan(60)=
bc
5+x
\tt \implies \sqrt{3} = \frac{5 + x}{bc}⟹
3
=
bc
5+x
\tt \implies bc = \frac{5 + x}{ \sqrt{3} } - - - (i)⟹bc=
3
5+x
−−−(i)
Now , in Δ BDC
\tt \implies\tan(30) = \frac{x}{bc}⟹tan(30)=
bc
x
\tt \implies \frac{1}{ \sqrt{3} } = \frac{x}{bc}⟹
3
1
=
bc
x
\tt \implies bc = \sqrt{3}x - - - (ii)⟹bc=
3
x−−−(ii)
From eq (i) and eq (ii) , we get
\tt \implies \frac{5 + x}{ \sqrt{3} } = \sqrt{3}x⟹
3
5+x
=
3
x
\tt \implies5 + x = 3x⟹5+x=3x
\tt \implies2x = 5⟹2x=5
\tt \implies x = \frac{5}{2}⟹x=
2
5
\tt \implies x = 2.5 \: \: m⟹x=2.5m
Therefore , the height of tower is 2.5 m
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