Question

The angle of elevation of top of building from foot of tower is 30degree and and angle of elevation of top of tower of building is 60degree if tower is 50 m height find the height of building

Answer 1

Answer:

Step-by-step explanation:

Let AB be the height of the building.

BC be the distance between the foots of the building and the tower.

Elevation is 30° and 60° from the tower and the building respectively.

ATP

In  right ΔBCD,

tan 60° = CD/BC

⇒ √3 = 50/BC

⇒ BC = 50/√3

also, In  right ΔABC,

tan 30° = AB/BC

⇒ 1/√3 = AB/BC

⇒ AB = 50/3

Thus, the height of the building is 50/3.

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:

$\sf In \: \Delta BDC$

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

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

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

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

___________________

$\sf In \: \Delta ABD,$

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

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

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

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

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

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