Question

The angle of elevation of the top of building from the foot of a tower is 30° & angle of elevation of top of tower from the foot of a building is 60°. If the tower is 50 mtr high. Find the height of the building.

Answer 1

here is ur answer:


height of tower = 50 m

let the height of building be AD m




In ∆ABC



$\tan(30) = \frac{p}{b} \\ \\ \frac{1}{ \sqrt{3} } = \frac{50}{ab} \\ \\ ab = 50 \sqrt{3}$





In ∆ ABD



$\tan(60) = \frac{p}{b} \\ \\ \sqrt{3} = \frac{ad}{50 \sqrt{3} } \\ \\ ad = 50 \times3 \\ \\ ad = 150$



height of building = 150 m


hope it helps you!
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Answer 2

Answer:

The angle of elevation of the top of a building from the foot of the tower is 30°.

The angle of elevation of the top of tower from the foot of the building is 60°.

Height of the tower is 50 m.

Let the height of the building be h.

Step-by-step explanation:

$\sf In \: \Delta BDC$

$:\implies \sf tan \: \theta = \dfrac{Perpendicular}{Base}$

$:\implies \sf tan \: 60^{\circ} = \dfrac{CD}{BD}$

$:\implies \sf \sqrt{3} = \dfrac{50}{BD}$

$:\implies \sf BD = \dfrac{50}{ \sqrt{3} } \: m$

___________________

$\sf In \: \Delta ABD,$

$\dashrightarrow\:\: \sf tan \: 30 ^{\circ} = \dfrac{AB}{BD}$

$\dashrightarrow\:\: \sf \dfrac{1}{ \sqrt{3} } = \dfrac{AB}{BD}$

$\dashrightarrow\:\: \sf \dfrac{1}{ \sqrt{3} } = \dfrac{h}{ \dfrac{50}{ \sqrt{3} } }$

$\dashrightarrow\:\: \sf h = \dfrac{ 50 }{ \sqrt{3} \times \sqrt{3} }$

$\dashrightarrow\:\: \underline{ \boxed{\sf Height= \dfrac{50}{3} \: m }}$

$\therefore\:\underline{\textsf{Height of the building is \textbf{50/3 meter}}}.$