## Height and Distance Questions and Answers Part-8

1. The angle of depression of a car parked on the road from the top of a 150 m high tower is 30°. The distance of the car from the tower (in metres) is
a) 50 $$\sqrt 3$$
b) 150 $$\sqrt 3$$
c) 100 $$\sqrt 3$$
d) 75

Explanation: Let AB be the tower of height 150 m
C is car and angle of depression is 30°
Therefore, ∠ACB = 30° (alternate angle)
In right - angled triangle ABC,
\eqalign{ & \frac{{BC}}{{AB}} = \cot {30^ \circ } \cr & \frac{{BC}}{{150}} = \sqrt 3 \cr & BC = 150\sqrt 3 \,m \cr}
That is, distance of the car from the tower is $$150\sqrt 3 \,m$$

2. If the altitude of the sun is at 60°, then the height of the vertical tower that will cast a shadow of length 30 m is
a) $$30\sqrt 3 \,m$$
b) $$15\,m$$
c) $$\frac{{30}}{{\sqrt 3 }}\,m$$
d) $$15\sqrt 2 \,m$$

Explanation: Let AB be tower and a point P distance of 30 m from its foot of the tower which form an angle of elevation pf the sun of 60°

\eqalign{ & {\text{Let height of tower }}AB = h \cr & {\text{Then in right }}\Delta APB, \cr & \tan \theta = \frac{{{\text{Perpendicular}}}}{{{\text{Base}}}} = \frac{{AB}}{{PB}} \cr & \Rightarrow \tan {60^ \circ } = \frac{h}{{30}} \cr & \sqrt 3 = \frac{h}{{30}} \cr & h = 30\sqrt 3 \cr}
Height of the tower $$= 30\sqrt 3 \,m$$

3. The tops of two poles of height 16 m and 10 m are connected by a wire of length l metres. If the wire makes an angle of 30° with the horizontal, then l =
a) 26
b) 16
c) 12
d) 10

Explanation: Let AB and CD are two poles AB = 10 m and CD = 16 m

AC is wire which makes an angle of 30° with the horizontal
Let BD = x, then AE = x
CE = CD - ED = CD - AB = 16 - 10 = 6m
\eqalign{ & {\text{Now}}\,{\text{in}}\,\Delta ACE \cr & \sin {30^ \circ } = \frac{{CE}}{{AC}} = \frac{6}{l} \cr & \frac{1}{2} = \frac{6}{l} \cr & l = 2 \times 6 = 12\,m \cr}

4. The height of a tower is 100 m. When the angle of elevation of the sun changes from 30° to 45°, the shadow of the tower becomes x metres less. The value of x is
a) $$100\,m$$
b) $$100\sqrt 3 \,m$$
c) $$100\left( {\sqrt 3 - 1} \right)\,m$$
d) $$\frac{{100}}{{\sqrt 3 }}\,m$$

Explanation: Let AB be tower and AB = 100 m and angles of elevation of A at C and D are 30° and 45° respectively and CD = x
Let BD = y

\eqalign{ & {\text{Now in right }}\Delta ADB, \cr & \tan \theta = \frac{{{\text{Perpendicular}}}}{{{\text{Base}}}} = \frac{{AB}}{{DB}} \cr & \tan {45^ \circ } = \frac{{100}}{y} \cr & \Rightarrow 1 = \frac{{100}}{y} \cr & \Rightarrow y = 100 \cr & {\text{Similarly in right }}\Delta ACB, \cr & \tan {30^ \circ } = \frac{{AB}}{{CB}} \cr & \frac{1}{{\sqrt 3 }} = \frac{{100}}{{y + x}} \cr & \frac{1}{{\sqrt 3 }} = \frac{{100}}{{100 + x}} \cr & 100 + x = 100\sqrt 3 \cr & x = 100\sqrt 3 - 100 \cr & x = 100\left( {\sqrt 3 - 1} \right)\,m \cr}

5. The angle of elevation of the top of a tower standing on a horizontal plane from a point A is α. After walking a distance 'd' towards the foot of the tower the angle of elevation is found to be β. The height of the tower is:
a) $$\frac{d}{{\cot \alpha + \cot \beta }}$$
b) $$\frac{d}{{\cot \alpha - \cot \beta }}$$
c) $$\frac{d}{{\tan \beta - \operatorname{tant} \alpha }}$$
d) $$\frac{d}{{\tan \beta + \operatorname{tant} \alpha }}$$

Explanation: Let AB be the tower and C is a point such that the angle of elevation of A is α.
After walking towards the foot B of the tower, at D the angle of elevation is β.
Let h be the height of the tower and DB = x Now in ΔACB,

\eqalign{ & \tan \theta = \frac{{{\text{Perpendicular}}}}{{{\text{Base}}}} = \frac{{AB}}{{CB}} \cr & \tan \alpha = \frac{h}{{d + x}} \cr & \Rightarrow d + x = \frac{h}{{\tan \alpha }} \cr & \Rightarrow d + x = h\cot \alpha \cr & \Rightarrow x = h\cot \alpha - d\,.......\left( {\text{i}} \right) \cr & {\text{Similarly in right }}\Delta ADB, \cr & \tan \beta = \frac{h}{x} \cr & \Rightarrow x = \frac{h}{{\tan \beta }} \cr & \Rightarrow x = h\cot \beta \,.........({\text{ii}}) \cr & {\text{From}}\,\left( {\text{i}} \right)\,{\text{and}}\,\left( {{\text{ii}}} \right) \cr & h\cot \alpha - d = h\cot \beta \cr & h\cot \alpha - h\cot \beta = d \cr & h\left( {\cot \alpha - \cot \beta } \right) = d \cr & h = \frac{d}{{\cot \alpha - \cot \beta }} \cr}
Height of the tower $$= \frac{d}{{\cot \alpha - \cot \beta }}$$

6. From a point P on a level ground, the angle of elevation of the top tower is 30º. If the tower is 200 m high, the distance of point P from the foot of the tower is:
a) 346 m
b) 400 m
c) 312 m
d) 298 m

Explanation:

\eqalign{ & \tan {30^ \circ } = \frac{{RQ}}{{PQ}} \cr & \frac{1}{{\sqrt 3 }} = \frac{{200}}{{PQ}} \cr & PQ = 200\sqrt 3 \cr & \,\,\,\,\,\,\,\,\,\, = 200 \times 1.73 \cr & \,\,\,\,\,\,\,\,\,\, = 346\,{\text{m}} \cr}

7. The angles of depression and elevation of the top of a wall 11 m high from top and bottom of a tree are 60° and 30° respectively. What is the height of the tree?
a) 22 m
b) 44 m
c) 33 m
d) None of these

Explanation:

Let DC be the wall, AB be the tree.
Given that ∠DBC = 30°, ∠DAE = 60°, DC = 11 m
\eqalign{ & \tan {30^ \circ } = \frac{{DC}}{{BC}} \cr & \frac{1}{{\sqrt 3 }} = \frac{{11}}{{BC}} \cr & BC = 11\sqrt 3 \,m \cr & AE = BC = 11\sqrt 3 \,m\,.....\left( 1 \right) \cr & \tan {60^ \circ } = \frac{{ED}}{{AE}} \cr}
$$\sqrt 3 = \frac{{ED}}{{11\sqrt 3 }}$$   [ Substituted value of AE from (1)]
\eqalign{ & ED = 11\sqrt 3 \times \sqrt 3 \cr & \,\,\,\,\,\,\,\,\,\, = 11 \times 3 \cr & \,\,\,\,\,\,\,\,\,\, = 33 \cr & {\text{Height}}\,{\text{of}}\,{\text{the}}\,{\text{tree}} \cr & = AB = EC = \left( {ED + DC} \right) \cr & = 33 + 11 \cr & = 44\,{\text{m}} \cr}

8. The angle of elevation of the top of the tower from a point on the ground is $${\sin ^{ - 1}}\left({\frac{3}{5}} \right).$$   If the point of observation is 20 meters away from the foot of the tower, what is the height of the tower?
a) 9 m
b) 18 m
c) 15 m
d) 12 m

Explanation: Consider a right-angled triangle PQR
Let QR = 3 and PR = 5 such that
$$\sin \theta = \frac{3}{5}\,\,\,\,\left[ {{\text{i}}{\text{.e}}{\text{.}},\theta = {{\sin }^{ - 1}}\left( {\frac{3}{5}} \right)} \right]$$
$$PQ = \sqrt {P{R^2} - Q{R^2}}$$     (∵ Pythagorean theorem)
\eqalign{ & = \sqrt {{5^2} - {3^2}} \cr & = 4 \cr}
$${\text{i}}{\text{.e}}{\text{.}},\,{\text{when}}\,\theta = {\sin ^{ - 1}}\left( {\frac{3}{5}} \right),$$     PQ : QR = 4 : 3 ......(eq : 1)

Let P be the point of observation and QR be the tower
$${\text{Given}}\,{\text{that}}\,\theta = {\sin ^{ - 1}}\left( {\frac{3}{5}} \right)$$     and PQ = 20 m
We know that PQ : QR = 4 : 3 (from eq : 1)
\eqalign{ & {\text{i}}{\text{.e}}{\text{.}},\,20:QR = 4:3 \cr & \Rightarrow 20 \times 3 = QR \times 4 \cr & \Rightarrow QR = 15\,{\text{m}} \cr}
Height of the tower = 15 m

9. A ladder 10 m long just reaches the top of a wall and makes an angle of 60° with the wall.Find the distance of the foot of the ladder from the wall $$\left( {\sqrt 3 = 1.73} \right)$$
a) 4.32 m
b) 17.3 m
c) 5 m
d) 8.65 m

Explanation:

Let BA be the ladder and AC be the wall as shown above.
Then the distance of the foot of the ladder from the wall = BC
Given that BA = 10 m, ∠ BAC = 60°
\eqalign{ & \sin {60^ \circ } = \frac{{BC}}{{BA}} \cr & \frac{{\sqrt 3 }}{2} = \frac{{BC}}{{10}} \cr & BC = 10 \times \frac{{\sqrt 3 }}{2} \cr & \,\,\,\,\,\,\,\,\,\, = 5 \times 1.73 \cr & \,\,\,\,\,\,\,\,\,\, = 8.65\,{\text{m}} \cr}

10. A man standing at a point P is watching the top of a tower, which makes an angle of elevation of 30º with the man's eye. The man walks some distance towards the tower to watch its top and the angle of the elevation becomes 45º. What is the distance between the base of the tower and the point P?
a) 9 units
b) $$3\sqrt 3$$ units
\eqalign{ & \tan {45^ \circ } = \frac{{SR}}{{QR}} \cr & \tan {30^ \circ } = \frac{{SR}}{{PR}} = \frac{{SR}}{{\left( {PQ + QR} \right)}} \cr}