^{0} when the boat is 80m from the tower. After 10 seconds, the angle of depression becomes 30^{0}. What is the speed of the boat? (Assume that the boat is running in still water).

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Two friends Ajay and Vijay formed sand castles of heights of height 8 cm and 15 cm respectively on the seashore. The distance between the two castles was 24 cm. Find the distance between the tops of two castles.

A.

24.5cm

B.

25cm

C.

31cm

D.

24cm

Answer with explanation

Answer: Option BExplanation

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When the sun’s altitude changes from 30° to 60°, the length of the shadow of a tower decreases by 70m. What is the height of the tower?

A.

140 m

B.

60.6 m

C.

35 m

D.

20.2 m

Answer with explanation

Answer: Option BExplanation

Let AD be the tower, BD be the initial shadow and CD be the final shadow.

Given that BC = 70 m, ABD = 30°, ACD = 60°,

Let CD = x, AD = h

From the right CDA,

tan60°=ADCD√3=hx⋯(eq:1)tan60°=ADCD3=hx⋯(eq:1)

From the right BDA,

tan30°=ADBD1√3=h70+x⋯(eq:2)tan30°=ADBD13=h70+x⋯(eq:2)

eq:1eq:2⇒√3(1√3)=(hx)(h70+x)⇒3=70+xx⇒2x=70⇒x=35eq:1eq:2⇒3(13)=(hx)(h70+x)⇒3=70+xx⇒2x=70⇒x=35

Substituting this value of x in eq:1, we have

√3=h35⇒h=35√3=35×1.73=60.55≈60.6

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^{0} when the boat is 80m from the tower. After 10 seconds, the angle of depression becomes 30^{0}. What is the speed of the boat? (Assume that the boat is running in still water).

A.

20 m/sec

B.

16 m/sec

C.

10 m/sec

D.

18 m/sec

Answer with explanation

Answer: Option BExplanation

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A man is standing on the deck of a ship, which is 10m above water level. He observes the angle of elevation of the top of a light house as 60^{0} and the angle of depression of the base of the lighthouse as 30^{0}. Find the height of the light house.

A.

45m

B.

30m

C.

38m

D.

40m

Answer with explanation

Answer: Option DExplanation

Workspace

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

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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

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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°

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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

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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

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