Two men are on the opposite side of a $40 m$ high tower. They measure the angles of elevation of the top of the tower $45^{o}$ and $30^{o}$ respectively the distance between men is :

40 (\sqrt{3} - 1) m

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40 \sqrt{3} m

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40 (\sqrt{3} + 1) m

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Solution

The correct option is B $40 (\sqrt{3} + 1) m$
Given,
Height of tower $h = 40 m$
for man A
$t a n 45 = \frac{h}{x}$

$⟹ x = \frac{40}{1} = 40 m$

For man $B$
$t a n 30 = \frac{h}{y}$
$\frac{1}{\sqrt{3}} = \frac{40}{y}$

$⟹ y = 40 \sqrt{3}$

Distance between two man $x + y = 40 + 40 \sqrt{3} = 40 (\sqrt{3} + 1) m$

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Q. Two men are on diametrically opposite side of a tower. They measure the angle of elevation of the top of the tower as

20^{0}

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24^{0}

respectively. If the height of the tower is

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[Hint.use tables to find tan

20^{0}

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Two towers are 40m apart. The angle of elevation of the top of the taller tower from the base and top of the shorter tower is $70^{\circ}$ and $40^{\circ}$ respectively. Find the height of the shorter tower. $[T a n 40^{\circ} = 0.83, T a n 70^{\circ} = 2.74]$

Two towers are 40m apart. The angle of elevation of the top of the taller tower from the base and top of the shorter tower is $70^{\circ}$ and $40^{\circ}$ respectively. The height of the shorter tower is $[T a n 40^{\circ} = 0.83, T a n 70^{\circ} = 2.74]$

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Two men are on the opposite side of a 40 m high tower. They measure the angles of elevation of the top of the tower 45o and 30o respectively the distance between men is :

The correct option is B 40(√3+1) mGiven,Height of tower h=40mfor man Atan45=hx⟹x=401=40mFor man Btan30=hy1√3=40y⟹y=40√3Distance between two man x+y=40+40√3=40(√3+1)m

Two men are on the opposite side of a $40 m$ high tower. They measure the angles of elevation of the top of the tower $45^{o}$ and $30^{o}$ respectively the distance between men is :