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- Question
- The angles of depression of two ships from the top of a light house are 45° and 30° towards east. if the ships are 200 m apart, find the height of the light house .
Options- A. 100 m
- B. 173 m
- C. 200 m
- D. 273 m
- Correct Answer
- 273 m
ExplanationLet us draw a figure below as per given question.
Let AB = h meter be the height of the tower and two ships are situated at D and C respectively; such that, CD = 200 m; ?ADC = 30° ?ACB = 45° and BC = x meter (say)
Now from right triangle ABC,
tan 45° = h/x ? 1 = h/x
? x = h
Again from right triangle ABD,
tan 30° = h/( 200 + x )
? 1/?3 = h/( 200 + x )
Since x = h , we will get.
? 1/?3 = h/(200 + h )
? (200 + h ) = h X ?3
? h X ?3 - h = 200
? h x 1.732 - h = 200
? h(1.732 - 1) = 200
? h = 200/0.732 = 273.2 m = 273 m - 1. Two posts are x meter apart and the height of one is double that of the other. If from the midpoint of the line joining their feet, an observer finds the angular elevations of their tops to be complementary, then the height (in meter) of the shorter post is :
Options- A. x 2 ?2
- B. x 4
- C. x ?2
- D. x ?2 Discuss
- 2. If r sin ? = 1, r cos ? = ? 3 then the value of ( ? 3 tan ? + 1 ) is :
Options- A. ?3
- B. 1 ?3
- C. 1
- D. 2 Discuss
- 3. The angle of elevation of a tower from a distance 50 m from its foot is 30°. The height of the tower is :
Options- A. 50 ?3 m
- B. 50 m ?3
- C. 75 ?3 m
- D. 75 m ?3 Discuss
- 4. From a 15 m high bridge on the river, the angle of depression of a boat is 30°. If the speed of the boat be 6 Km/h, then the time taken by the boat to reach exactly below the bridge will be :
Options- A. 9 ?3 second
- B. 19 / ?3 second
- C. 3 ?3 second
- D. None of these Discuss
- 5. The top of a 15 m high tower makes an angle of elevation of 60° with the bottom of an electric pole and an angle of elevation of 30° with the top of the pole. What is the height of the electric pole?
Options- A. 5 m
- B. 8 m
- C. 10 m
- D. 12 m Discuss
- 6. At a point on a level plane a tower subtends an angle ? and a flag-staff a ft. in length at the top of the tower subtends an angle ? . The height of the tower is :
Options- A. aSin?Cos?/Cos(? + ?)
- B. aSin?Cos(? + ?)/Sin?
- C. aCos(? + ?)/Sin?Sin?
- D. None of these Discuss
- 7. A 6 ft-tall man finds that the angle of elevation of the top of a 24 ft-high pillar and the angle of depression its base are complementary angles. Then the distance between pillar and man is?
Options- A. 2?3 ft
- B. 4?3 ft
- C. 6?3 ft
- D. 8?3 ft Discuss
- 8. Two ships leave a port at the same instant. One sails at 30km/hr in the direction N 32° E while the other sails at 20 km/hr in the direction S 58° E. After two hours the ships are distant from each other by.
Options- A. 15?6 Km
- B. 36.5 Km
- C. 20?13 Km
- D. 100 Km Discuss
- 9. ABC is a triangular park with AB = AC = 100 meters. A clock tower is situated at the mid-point of BC. The angles of elevation of the top of the tower of A and B are cot -1 (3.2) and cosec -1(2.6) respectively. The height of the tower in meters is?
Options- A. 25/2
- B. 25
- C. 50
- D. None of these Discuss
- 10. ABCD is a square plot. The angle of elevation of the top of a pole standing at D from A or C is 30° and that from B is ? then Tan ? equal to?
Options- A. ?6
- B. 1/?6
- C. ?3/?2
- D. ?2/?3 Discuss
Height and Distance problems
Search Results
Correct Answer: x 2 ?2
Explanation:
Correct Answer: 2
Explanation:
Correct Answer: 50 m ?3
Explanation:
Correct Answer: 9 ?3 second
Explanation:
Correct Answer: 10 m
Explanation:
Correct Answer: aSin?Cos(? + ?)/Sin?
Explanation:
Let us draw a figure as per given question.
Let OP be the tower of height h (say) and PQ the flag-staff of height a, such that ?OAP = ? and ?PAQ = ?
In ?OAP use the formula
Tan? = P/B = Perpendicular distance / Base distance
? Tan? = OP/OA
? OA = OP Cot?
Put the value of OP in above equation, we will get
? OA = hCot? ................ (1)
Now from triangle ?OAQ,
? Tan(? + ?) = OQ/OA
? OA = OQ Cot(? + ?)
? OA = (h + a) Cot(? + ?) .............(2)
from equation (1) and (2), We will get
? hCot? = (h + a) Cot(? + ?)
? Cot? / Cot(? + ?) = (h + a) / h
? Cot? / Cot(? + ?) = h / h + a / h
? Cot? / Cot(? + ?) = 1 + a / h
? (Cot? / Cot(? + ?)) - 1 = a / h
? (Cot? - Cot(? + ?) ) / Cot(? + ?) = a / h
? h = aCot (? + ?) / Cot? - Cot(? + ?)
After converting Cot into Sin and Cos, We will get
? h = aSin?Cos(? + ?)/Sin?
Correct Answer: 6?3 ft
Explanation:
Let us draw a figure below from the given question.
Let AB = 24 ft and CD = 6 ft be the height of the pillar and man respectively.
Here ? ACE = ? ? CBD = 90° - ?
Now from right triangle CDB,
BD = 6 cot (90° - ?) = 6 tan ?
? tan ? = BD/6 ...............(i)
From right triangle ACE,
tan ? =18/EC = 18/BD ...............(ii) (? EC = BD)
Now from (i) and (ii) we get,
BD/6 = 18/BD
? BD2 = 18 x 6
? BD = 6?3 ft.
Hence, distance between pillar and man = 6?3 ft
Correct Answer: 20?13 Km
Explanation:
Let us draw a figure below as per given question.
Let two ships started from point O at the speed of 30 km/hr and 20 km/hr respectively, after two hours they reach at points A and B.
Now, ?NOA = 32° and ?SOB = 58° ,
Then, ?AOB = 180° - (32° + 58°) = 90°
Since ?AOB is a right triangle in which OA = 2 x 30 = 60 km and OB = 2 x 20 = 40 km
Since, AB = ?OA2 + OB2
= ?(60)2 + (40)2
= ?5200
= 20?13
Correct Answer: 25
Explanation:
Let us draw a figure below as per given question.
Given ? = cot-1 (3.2) and ? = cosec-1(2.6)
In triangle ?PAD, AD = h cot ?
In triangle ?PBD, BD = h cot ?
In triangle ? ABD,
AB2 = AD2 + BD2
= h2(cot2? + cot2 ? )
? 1002 = h2{ cot2 ? + (cosec2 ? - 1) }
? 1002= h2 { (3.2)2 + (2.6)2 - 1} = 16h2
? h = 25 m.
Correct Answer: 1/?6
Explanation:
Let a be the length of a side of square plot ABCD and h, the height of the pole standing at D. Since elevations of P from A or C is 30° and that from B is ?,
? In triangle ? PCD, tan 30° = h/a
i.e h/a = 1/?3 .............. (1)
And in triangle ? PBD,
Since PD = h and BD = ?(AB2 + AD)2 = a?2.
Put the value of PD and BD , we will get
tan ? = PD/BD = h/(a?2)
put the value of h/a from equation (1)
tan ? = 1/?2 x 1/?3
tan ? = 1/?6
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