Height and Distance problems
- 1. The angles of elevation of the top of a verticle tower from two points, distance a and b (a > b) from the base and in the same straight line with it are complementary. Then the height of the tower is?
Options- A. ?(ab)
- B. ?(a2 + b2)
- C. ?(a2 - b2)
- D. ?a(a - b) Discuss
Correct Answer: ?(ab)
Explanation:
Let CD = h unit be the height of the tower and A and B be the two points on the ground, such that DA = a; DB = b;
? DAC = ? and ?DBC = 90° - ?
From right triangle ADC, CD = h = a tan ? ...(i)
From right triangle BDC, CD = h = b tan (90° - ? ) = b cot ? .......(ii)
Multiplying equations (i) and (ii), we get
h2 = a tan ? X b cot?
Hence, h = ?ab
- 2. The angle of elevation of the top of a tower standing on a horizontal plane from a point A is ?. After walking a distance a towards the foot of the tower, the angle of elevation is found to be ?. The height of the tower is :
Options- A. a Sin ? Sin ?/ Sin(? - ?)
- B. a Sin ? Sin ?/Sin(? - ?)
- C. a Sin( ? - ? ) Sin ? Sin ?
- D. a Sin(? - ?)/Sin ?Sin ? Discuss
Correct Answer: a Sin ? Sin ?/ Sin(? - ?)
Explanation:
Let OP be the tower of height h (say) and A and B be the two positions on the horizontal line through O, such that
?OAP = ?, ?OBP = ? and OB = x
In ?OBP, Use the trigonometry formula
Tan? = P/B = Perpendicular distance / Base distance
Tan? = OP/OB
? OB = OP/Tan?
? OB = OP Cot?
Put the value of OB and OP , We will get
x = h Cot ?...............(1)
In ?OAP, Similarly
Tan? = OP/OA
? OA = OP/ Tan?
? OA = OP Cot ?
Put the value of OA and OP
? a + x = h Cot ?
? x = h Cot ? - a ............(2)
From equation (1) and (2)
? h Cot ? = h Cot ? - a
? a = h Cot ? - h Cot ?
? a = h (Cot ? - Cot ?)
? a = h (Cos ?/ Sin ? - Cos ? / Sin ? )
? a = h( (Cos ? Sin ? - Cos ? Sin ? ) /Sin ? Sin ? )
? a = h( Sin(? - ?) / Sin ? Sin ?)
? h = a Sin ? Sin ?/ Sin(? - ?)
- 3. A man on the top of a vertical towers observes a car moving at a uniform speed coming directly towards it. if it takes 12 minute for the angle of depression to change from 30° to 45°, how soon after this will the car reach the tower?
Options- A. 14 min 20 sec.
- B. 15 min 22 sec.
- C. 16 min.
- D. 16 min 23 sec. Discuss
Correct Answer: 16 min 23 sec.
Explanation:
Let us draw a figure below from given question.
Let AB = h meter be the height of the tower B and C are two points such that ?ACB = 30° ?ADB = 45° and CD = x meter (say)
From right triangle ABD,
tan 45° = h/BD
? BD = h meter;
Again from right triangle ABC
tan 30° = h/(h + x )
? h + x = ?3 h
? x = (1.73 - 1)h = 0.73h
Now, 0.73h meter covered in 12 min
Hence, h meter covered in 12/0.73 = 1200/73 min = 16 min 23 sec .
- 4. An observer measures angles of elevation of two tower of equal height from a point between the towers. If the angles of elevation are 60° and 30° and distance of nearer tower is 100 m then the height of each tower and the distance between the towers, respectively are
Options- A. 100/?3 m and 300 m
- B. 100/?3 m and 400 m
- C. 100?3 m and 300 m
- D. 100?3 m and 400 m Discuss
Correct Answer: 100/?3 m and 400 m
Explanation:
Let us draw a figure below as per given question.
Let AB = CD = h meter be the heights of the towers. E is a point such that DE = 100 meter;
?CED = 60° and ?AEB = 30°
Now, BE = x meter (say)
From right triangle CDE.
h = 100 tan 60°
? h = 100?3 meter
From right triangle ABE,
x = h cot 30° put the value of h, we will get
x = 100?3 X ?3
x = 100 X 3 = 300 meters
Distance between the tower = DE + EB = 100 + 300 = 400 meters
Height of the tower = h = 100?3 meter
- 5. A man 2 m high, walks at a uniform speed of 6 m/min away from a lamp post, 5 m high. Find the rate at which the length of his shadow increases .
Options- A. 4 m/min
- B. 8 m/min
- C. 9m/min
- D. 14 m/min Discuss
Correct Answer: 4 m/min
Explanation:
Let us draw a figure as per given question.
Let AB = 5 meter and CD = 2 meter be the heights of a lamp post and the men respectively.
At any time t BD = x meter and shadow of man ED = y meter.
Then, dx/dt = 6 m/min
Now, right triangles ABE and CDE are similar, then
AB/CD = BE/DE
? 5/2 = (x + y)/y
? 5y = 2x + 2y
? 5y - 2y = 2x
? 3y = 2x
? 3 dy/dt = 2 dx/dt
? 3 dy/dt = 2 x 6
? dy/dt = 2 x 6 / 3
? dy/dt = 2 x 2 = 4
Hence, length of his shadow increase at the rate of 4 m/min.
- 6. The angle of elevation of moon when the length of the shadow of a pole is equal to its height, is:
Options- A. 30°
- B. 45°
- C. 60°
- D. None of these Discuss
Correct Answer: 45°
Explanation:
- 7. A pole being broken by the wind, the top struck the ground at an angle of 30° and at a distance of 21 m from the foot of the pole. Find out the total height of the pole.
Options- A. 21 m
- B. 21 ?3 m
- C. 21/ ?3
- D. None of these Discuss
Correct Answer: 21 ?3 m
Explanation:
- 8. The shadow of a tower standing on a level plane is found to be 50 m longer when the sun's altitude is 30° than when it is 60°. Find the height of the tower.
Options- A. 20 ?3 m
- B. 25/ ?3 m
- C. 25 ?3 m
- D. 20 ?3 m Discuss
Correct Answer: 25 ?3 m
Explanation:
- 9. A 20 m high electric pole stands upright on the ground with the help of steel wire to its top and affixed on the ground. If the steel wire makes 60° with the horizontal ground, then find out the length of the steel wire.
Options- A. 40/ ?3 m
- B. 40 ?3 m
- C. 20/ ?3m
- D. 20 ?3m Discuss
Correct Answer: 40/ ?3 m
Explanation:
- 10. When the length of the shadow of a pole is equal to the height of the pole, then the elevation of source of light is :
Options- A. 30°
- B. 45°
- C. 60°
- D. 75° Discuss
Correct Answer: 45°
Explanation:
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