# Height and Distance Questions Answers

1) Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 30 degree and 45 degree respectively. If the lighthouse is 100 m high, the distance between the two ships i

- 1) 276 metre
- 2) 273 metre
- 3) 270 metre
- 4) 263 metre

**Ans.**B

**Explanation :**

Let AB be the lighthouse and C and D be the positions of the ships. \begin{aligned} \text{AB = 100m}, \angle{ACB}=30^{\circ}, \\angle{ADB}=45^{\circ}\\frac{AB}{AC} = tan&30{\circ} = \frac{1}{\sqrt{3}} \=> AC = AB*\sqrt{3} = 100\sqrt{3}m\ \frac{AB}{AD} = tan&45^{\circ} = 1 \=> AB = AD = 100m\ CD = AC+AD\= (100\sqrt{3}+100)&m \= 100(\sqrt{3}+1)&m\= 100*2.73&m = 273m \end{aligned}

2) The angle of elevation of the sun, when the length of the shadow of a tree is \begin{aligned}\sqrt{3}\end{aligned} times the height of tree, is

- 1) 30 degree
- 2) 45 degree
- 3) 60 degree
- 4) 9 degree

**Ans.**A

**Explanation :**

Let AB be the tree and AC be its shadow, So as per question, shadow of a tree is \begin{aligned}\sqrt{3}\end{aligned} times the height of tree. Let h be the height of the tree, then \begin{aligned} \frac{AB}{AC} = tan\theta \=> \frac{h}{\sqrt{3}h} = tan\theta \=> tan\theta = \frac{1}{\sqrt{3}} \=> \theta = 30^{\circ} \end{aligned}

3) From a point C on a level ground, the angle of elevation of the top of a tower is 30 degree. If the tower is 100 meter high, find the distance from point C to the foot of the towe

- 1) 170 meter
- 2) 172 meter
- 3) 173 meter
- 4) 167 meter

**Ans.**C

**Explanation :**

Let AB be the tower. \begin{aligned} then \angle{ACB}= 30^{\circ} \AB = 100&meter \\frac{AB}{AC} = tan&30^{\circ} \=>\frac{100}{AC} = \frac{1}{\sqrt{3}} \=> AC = \sqrt{3}*100 \=> AC = 1.73*100 \=> AC = 173 meter \end{aligned} Please always remember the value of square root 3 is 1.73, and value of square root 2 is 1.41. This will be very helpful while solving height and distance questions and saving your time.

4) A man standing at a point C is watching the top of a tower, which makes an angle of elevation of 30 degree. The man walks some distance towards the tower to watch its top and the angle of elevation become 60 degree. What is the distance between the base of the tower and point C

- 1) \begin{aligned} 4\sqrt{3} meter \end{aligned}
- 2) \begin{aligned} 2\sqrt{3} meter \end{aligned}
- 3) \begin{aligned} \sqrt{3} meter \end{aligned}
- 4) Data is inadequate

**Ans.**D

**Explanation :**

Please refer to the image. One of the AB, AD and CD must be given. So data is inadequate.

5) A man is watching form the top of the tower a boat speeding away from the tower. The boat makes the angle of depression of 45 degree with the man's eye when at a distance of 60 metres from the tower. After 5 seconds the angle of depression becomes 30 degree. What is the approximate speed of the boat, assuming that it is running in still water

- 1) 22 Km/Hr
- 2) 28 Km/Hr
- 3) 32 Km/Hr
- 4) 36 Km/Hr

**Ans.**C

**Explanation :**

Let AB be the tower and C and D be the positions of the boat. \begin{aligned} then \angle{ACB}=45^{\circ}, \angle{ADB}=30^{\circ}, \AC = 60&m \Let& AB = h \=>\frac{AB}{AC} = tan&45^{\circ} = 1 \=> AB = AC = 60&m \ and, \frac{AB}{AD} = tan&30^{\circ} = \frac{1}{\sqrt{3}} \=> AD = AB*\sqrt{3} = 60\sqrt{3}&m \ => CD = AD-AC = 60\sqrt{3}-60\=> CD = 60(\sqrt{3}-1) m \\text{We Know Speed = }\frac{Distance}{Time} \=> Speed = \left[\frac{60(\sqrt{3}-1)}{5} \right]m/sec = 12*0.73&m/sec \ \text{Please note answer we need in Km/Hr} \=> Speed = 12*0.73*\frac{18}{5} Km/Hr \=> Speed = 31.5 & Km/hr, \\text{Which is approx 32 Km/Hr} \end{aligned}

6) The Top of a 15 metre high tower makes an angle of elevation of 60 degree with the bottom of an electric pole and angle of elevation of 30 degree with the top of pole. Find the height of the electric pol

- 1) 7 metre
- 2) 8 metre
- 3) 9 metre
- 4) 10 metre

**Ans.**D

**Explanation :**

Let AB be the tower and CD be the electric pole. \begin{aligned} \angle{ACB}=60^{\circ}, \angle{EDB}=30^{\circ}, \AB = 15&m \Let & CD = h, \text{ then}, \BE = AB-AE = AB - AE = 15-h\\frac{AB}{AC} = tan 60^{\circ} = \sqrt{3} \=> AC = \frac{AB}{\sqrt{3}} \=> AC = \frac{15}{\sqrt{3}} \and, & \frac{BE}{DE} = tan 30^{\circ} = \frac{1}{\sqrt{3}} \=> DE = (BE*\sqrt{3}) \= \sqrt{3}(15-h) \ Now, & AC = DE \=> \frac{15}{\sqrt{3}} = \sqrt{3}(15-h) \=> 3h = 45-15 \=> h = \frac{30}{3} = 10 m \end{aligned}