Question

The angles of depression of the top and the bottom of an 8m tall building from the top of a multi-storeyed building are 30° and 45° , respectively. Find the height of the multi-storeyed building and the distance between the two buildings.

Answer 1

Given: The angles of depression of the top and the bottom of an 8 m tall building from the top of a multi-storeyed building are 30° and 45°.

To find: The height of the multi-storeyed building and the distance between the two buildings.

Answer:

(Diagram for reference attached below.)

In Δ ACD,

$\tt tan\ {45}^{\circ}\ =\ \dfrac{AC}{DC}\\\\\\\sf [tan\ {45}^{\circ}\ =\ 1]\\\\\\\tt 1\ =\ \dfrac{x\ +\ 8}{DC}\\\\\\DC\ =\ x\ +\ 8$

In Δ ABE,

$\tt tan\ {30}^{\circ}\ =\ \dfrac{AB}{BE}\\\\\\\sf [tan\ {30}^{\circ}\ =\ \dfrac{1}{\sqrt{3}}]\\\\\\\tt \dfrac{1}{\sqrt{3}}\ =\ \dfrac{x}{x\ +\ 8}\\\\\\x\ +\ 8\ =\ \sqrt{3}x\\\\\\8\ =\ \sqrt{3}x\ -\ x\\\\\\8\ =\ x(\sqrt{3}\ -\ 1)\\\\\\\dfrac{8}{\sqrt{3}\ -\ 1}\ =\ x\\\\\\\dfrac{8\ \times\ (\sqrt{3}\ +\ 1)}{(\sqrt{3}\ -\ 1)(\sqrt{3}\ +\ 1)}\ =\ x\\\\\\\dfrac{8\ \times\ (\sqrt{3}\ +\ 1)}{2}\ =\ x\\\\\\4(\sqrt{3}\ +\ 1)\ =\ x$

Therefore, the height of the multi-storeyed building [AC] = [4(√3 + 1) + 8] m.

And the distance between the two buildings is  [4(√3 + 1) + 8] m as well.

Answer 2

CRM

PLEASE CHECK YR ANSWER CAREFULLY

HOPE IT HELPS TO U

PLEASE MARK ME AS BRAINLIST