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- Question
- 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. a cos ( ? + ? ) / sin ? sin ?
- D. None of these
- Correct Answer
- None of these
ExplanationLet OP be the tower of height h (say) and PQ the flag-staff of height a, such that
?OAP = ? and ?PAQ = ?
In ?OAP and ?OAQ
OA = OP cot ? = h cot?
and OA = OQ cot ( ? + ? ) = (h + a) cot ( ? + ? )
? h cot? = (h + a) cot ( ? + ? )
? h = a cot ( ? + ? ) / cot? - cot ( ? + ? ) - 1. The height of the center of the round balloon of radius r, which subtend an angle ? at the eye of an observer and the elevation of whose center from the eye is ?, is given by?
Options- A. r Sin ? Sin ?
- B. r Cosec ?/2 Sin ?
- C. r Cosec ? Sin ?
- D. None of these Discuss
- 2. 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
- 3. 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
- 4. 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
- 5. 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
- 6. A tower stands on a horizontal plane. A man on the ground 100 m from the base of the tower finds the angle of elevation of the top of the tower to be 30°. What is the height of the tower?
Options- A. 100 m
- B. 100 ?3
- C. 100/ ?3
- D. None of these Discuss
- 7. The upper part of a tree broken by the wind makes an angle of 30° with the ground and the distance from the root to the point where the top of the tree touches the ground is 10 m. What was the height of the tree?
Options- A. 10?3
- B. 10/ ?3
- C. 20 ?3
- D. None of these Discuss
- 8. In a rectangle, if the angle between a diagonal and a side is 30° and the length of diagonal is 6 cm, the area of the rectangle is :
Options- A. (a) 9 cm2
- B. 9 ?3 cm2
- C. 27 cm2
- D. 36 cm2 Discuss
- 9. From the top of a lighthouse, 50 m above the sea, the angle of depression of an incoming boat is 30°. How far is the boat from the lighthouse?
Options- A. 25 ?3 m
- B. 25 / ?3 m
- C. 50 ?3 m
- D. 50 / ?3 m Discuss
- 10. On the level ground, the angle of elevation of the top of the tower is 30°. On moving 20 m nearer, the angle of elevation is 60°. The height of the tower is :
Options- A. 20 ?3 m
- B. 10 ?3 m
- C. 10 ( ?3 - 1 ) m
- D. None of these Discuss
Height and Distance problems
Search Results
Correct Answer: r Cosec ?/2 Sin ?
Explanation:
Let O be the centre of the balloon of radius r which subtend an angle ? at the eye of an observer at E .
If EA and EB are the tangents to the ballon,
then ? OEA = ? OEB = ?/2
In triangle ?OAE, Sin ?/2 = OA/OE
? OE = r cosec 1/2 ?
In ?OEL, height of the center of the balloon = h = OE sin ? = r Cosec ?/2 Sin ?.
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
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: 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: 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: 100/ ?3
Explanation:
Correct Answer: 10?3
Explanation:
Correct Answer: 9 ?3 cm2
Explanation:
Correct Answer: 50 ?3 m
Explanation:
Correct Answer: 10 ?3 m
Explanation:
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