- « Previous Page
- 1
- …
- 4
- 5
- 6

A Learning Portal from Recruitment India

Home » Aptitude » Height & Distance » Page 6

Consider a regular hexagon ABCDEF. There are towers placed at B and D. The angle of elevation from A to the tower at B is 30 degrees, and to the top of the tower at D is 45 degrees. What is the ratio of the heights of towers at B and D?

A.

1 : 2

B.

1 : 3–√

C.

3 : 43–√

D.

1 : 2√3

Answer with explanation

Answer: Option DExplanation

Workspace

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 30º and 45º respectively. If the lighthouse is 100 m high, the distance between the two ships is

A.

273 m

B.

300 m

C.

200 m

D.

173 m

Answer with explanation

Answer: Option AExplanation

Let BD be the lighthouse and A and C be the positions of the ships.

Then, BD = 100 m, BAD = 30° , BCD = 45°

tan30°=BDBA⇒1√3=100BA⇒BA=100√3tan30°=BDBA⇒13=100BA⇒BA=1003

tan45°=BDBC⇒1=100BC⇒BC=100tan45°=BDBC⇒1=100BC⇒BC=100

Distance between the two ships

= AC = BA + BC

=100√3+100=100(√3+1)=1003+100=100(3+1)

= 100(1.73+1) = 100 × 2.73 = 273 m

Workspace

A balloon leaves the earth at a point A and rises vertically at uniform speed. At the end of 2 minutes, John finds the angular elevation of the balloon as 60°. If the point at which John is standing is 150 m away from point A, what is the speed of the balloon?

A.

3.87 meter/sec

B.

0.63 meter/sec

C.

0.72 meter/sec

D.

2.16 meter/sec

Answer with explanation

Answer: Option DExplanation

Let C be the position of John. Let A be the position at which balloon leaves the earth and B be the position of the balloon after 2 minutes.

Given that CA = 150 m, BCA = 60°

Workspace

Find the angle of elevation of the sun when the shadow of a pole of 18 m height is 6 √ 3 m long?

A.

90°

B.

80°

C.

70°

D.

60°

Answer with explanation

Answer: Option DExplanation

Let RQ be the pole and PQ be the shadow

Given that RQ = 18 m and PQ =6√3 m

Let the angle of elevation, RPQ = θ

From the right PQR,

tanθ=RQ/PQ

=18/(6√3)

=3/√3

=√3

⇒θ=tan−1(√3)=60°

Workspace

Jack takes 20 minutes to jog around the race course one time, and 25 minutes to jog around a second time. What is his average speed in miles per hour for the whole jog if the course is 3 miles long?

A.

10

B.

9

C.

8

D.

6

Answer with explanation

Answer: Option CExplanation

Average speed = total distance / total time

Total distance covered = 6 miles; total time = 45 minutes = 0.75 hours

Average speed = 6/ 0.75 = 8 miles/hour

Workspace

The top of a 15 metre high tower makes an angle of elevation of 60° with the bottom of an electronic pole and angle of elevation of 30° with the top of the pole. What is the height of the electric pole?

A.

12 m

B.

10 m

C.

8 m

D.

6 m

Answer with explanation

Answer: Option BExplanation

Consider the diagram shown above. AC represents the tower and DE represents the pole.

Given that AC = 15 m, ADB = 30°, AEC = 60°

Let DE = h

Then, BC = DE = h,

AB = (15-h) (∵ AC=15 and BC = h),

BD = CE

Workspace

Two persons are on either side of a tower of height 50 m. The persons observers the top of the tower at an angle of elevation of 30° and 60°. If a car crosses these two persons in 10 seconds, what is

A.

21√3 km/hr

B.

24√3 km/hr

C.

17√3 km/hr

D.

22√3 km/hr

Answer with explanation

Answer: Option BExplanation

Let BD be the tower and A and C be the positions of the persons.

Given that BD = 50 m, BAD = 30°, BCD = 60°

From the right ABD,

tan 30° = BD/BA

⇒1/√3=50/BA

⇒BA=50√3

From the right CBD,

tan 60° = BD/BC

⇒√3=50/BC

⇒BC=50/√3

=(50×√3)/(√3×√3)

=(50√3)/3

Distance between the two persons

= AC = BA + BC

=50√3+((50√3)/3)

=√3(50+(50/3))

=(200√3)/3 m

i.e., the distance travelled by car in 10 seconds =(200√3)/3 m

Speed of the car =Distance/Time

=((200√3/3)/10)

=(20√3)/3 meter/second

=(20√3)/3×(18/5) km/hr

=24√3 km/hr

Workspace

The angle of elevation of the top of a tower from a certain point is 30°. If the observer moves 20 m towards the tower, the angle of elevation of the top of the tower increases by 15°. The height of the tower is

A.

21.9m

B.

30m

C.

17.3m

D.

27.3m

Answer with explanation

Answer: Option DExplanation

Workspace

A vertical toy 18 cm long casts a shadow 8 cm long on the ground. At the same time a pole casts a shadow 48 m. long on the ground. Then find the height of the pole ?

A.

1080 m

B.

108 m

C.

180 m

D.

118 m

Answer with explanation

Answer: Option BExplanation

We know the rule that,

At the particular time for all object, the ratio of height and shadow are same.

Let the height of the pole be ‘H’

Then, 18/8 = H/48

=> H = 108 m.

Workspace

A vertical post 15 ft. high is broken at a certain height and its upper part, not completely separated meets the ground at an angle of 30°. Find the height at which the post is broken

A.

5 ft.

B.

5√3 ft.

C.

10 ft.

D.

15√3 ft.

Answer with explanation

Answer: Option AExplanation

Workspace

- « Previous Page
- 1
- …
- 4
- 5
- 6

Correct Answer 👍

Wrong Answer 👎

RecruitmentIndia.in is Blog where we will update the information by exploring various online and offline sources of information. Our aim is to provide the latest Education related news as fast as possible to the students for free of cost.

Exams.Recruitmentindia.in is a Preparation portal where you can prepare all the competitive related questions like Aptitude, Reasoning, English Questions and Current affairs for free of cost

You can Contact us at :