Question

From the top of a 60 meter height building, the angles of
depression to the top and foot of a cell tower are 30 and
60° respectively. Find the height of the cell tower.

Answer 1

Answer:

height of the cell tower is 20m

Answer 2

Height of the cell tower is $40\ cm$

Step-by-step explanation:

Given,

Height of the building $60\ m$ and the angles of  depression to the top and foot of a cell tower are $30\ degree\ and\ 60\ degree$

From figure,

$AD=$Height of the building.

$BC=$Height of the cell tower.

$BE=CD$,$BC=DE$,$\angle ABE=30\ degree$ and $\angle ACD=60\ degree$

From $\triangle ACD$,

$tan60=\frac{AD}{CD}$

⇒$CD=\frac{60}{\sqrt{3}} =20\sqrt{3}$ $cm$

∴$BE=CD=20\sqrt{3}\ cm$

From $\triangle ABE$,

$tan30=\frac{AE}{BE}$

⇒$AE=20\sqrt{3}\times \frac{1}{\sqrt{3}}$

⇒$AE=20\ cm$

∴ $DE=AD-AE$

⇒$DE=60-20$

⇒$DE=40\ cm$

So, Height of the cell tower is $40\ cm$