Example 6: 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°, respectively. Find the height
202
MATICS
golul
repr
opp
AB
P
300450
OP
of the multi-storeyed building and the distance between the two buildings.
Solution : In Fig. 9.9, PC denotes the multi-
Q
storyed building and AB denotes the 8 m tall
building. We are interested to determine the
height of the multi-storeyed building, i.e., PC
and the distance between the two buildings,
i.e., AC
of
in
o
B
D
Look at the figure carefully. Observe that
PB is a transversal to the parallel lines PQ
and BD. Therefore, 2 QPB and Z PBD are
alternate angles, and so are equal.
A
С
T
Answers
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,
\begin{gathered}\tt tan\ {45}^{\circ}\ =\ \dfrac{AC}{DC}\\\\\\\sf [tan\ {45}^{\circ}\ =\ 1]\\\\\\\tt 1\ =\ \dfrac{x\ +\ 8}{DC}\\\\\\DC\ =\ x\ +\ 8\end{gathered}
tan 45
∘
=
DC
AC
[tan 45
∘
= 1]
1 =
DC
x + 8
DC = x + 8
In Δ ABE,
\begin{gathered}\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\end{gathered}
tan 30
∘
=
BE
AB
[tan 30
∘
=
3
1
]
3
1
=
x + 8
x
x + 8 =
3
x
8 =
3
x − x
8 = x(
3
− 1)
3
− 1
8
= x
(
3
− 1)(
3
+ 1)
8 × (
3
+ 1)
= x
2
8 × (
3
+ 1)
= x
4(
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.