^{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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A man 6ft tall cats a shadow 4ft long. At the same time when a flag pole casts a shadow 50 ft long. The height of the flag pole is

A.

15 ft

B.

75 ft

C.

60 ft

D.

70 ft

Answer with explanation

Answer: Option BExplanation

**Let x represent how tall the flagpole is
Question Sets Up the Proportion
**

** x/10=6/4
Cross Multiplying
4x = 60
x = 15ft tall**

** **

Workspace

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 600 and the angle of depression of the base of lighthouse as 300. Find the height of the light house.

A.

30m

B.

40m

C.

45m

D.

38m

Answer with explanation

Answer: Option BExplanation

from A is 60 °

Then ∠EAD= 60° &

∠CAE= ∠BCA= 30°. (Alternate ANGLES)

Let AD = BC = x m & DE= h m

In ∆ ADE

tan 60° = Perpendicular / base = DE/AD

√3= h/x [tan 60° = √3]

h = √3x……..(1)

In ∆ ABC

tan 30° = AB /BC

[ tan30° = 1/√3]

1/√3 = 10/x

x= 10√3 m.. …………..(2)

Substitute the value of x from equation (2) in equation (1), we have

h = √3x

h= √3× 10√3= 10 × 3= 30 m

h = 30 m

The height of the hill is CE= CD+ DE= 10 +30= 40 m

Hence, the height of the hill is 40 m & the Distance of the hill from the ship is 10√3 m.

Read more on Brainly.in – https://brainly.in/question/1001119#readmore

Workspace

A.

32 kmph

B.

36 kmph

C.

38 kmph

D.

40 kmph

Answer with explanation

Answer: Option AExplanation

Let AB be the tower and C and D be the two positions of the boats.

Then,

∠ACB = 45∘,∠ACB = 45∘,

∠ADB=30∘∠ADB=30∘

and AC = 60 m

Let, AB = h

Then,ABAC=tan45∘=1⇒AB=AC⇒h=60mAndABAD=tan30∘=13–√⇒AD=(AB×3–√)=603–√m∴CD=(AD−AC)=60(3–√−1)mHence, required speed = [60(3–√−1)5]m/s = (12×0.73)m/s=(12×0.73×185)km/hr=31.5km/hr ≈ 32km/hr

Workspace

From a point P on a level ground, the angle of elevation of the top tower is 30°. If the tower is 100 m high, the distance of point P from the foot of the tower is:

A.

149 m

B.

156 m

C.

173 m

D.

200 m

Answer with explanation

Answer: Option CExplanation

Let AB be the tower.

Then, APB = 30° and AB = 100 m.

AB | = tan 30° = | 1 |

AP | 3 |

AP | = (AB x 3) m |

= 1003 m | |

= (100 x 1.73) m | |

= 173 m. |

Workspace

The angle of elevation of the sun, when the length of the shadow of a tree 3 times the height of the tree, is:

A.

30°

B.

45°

C.

60°

D.

90°

Answer with explanation

Answer: Option AExplanation

Let AB be the tree and AC be its shadow.

Let ACB = .

Then, | AC | = | 3 cot = 3 |

AB |

= 30°.

Workspace

The angle of elevation of a ladder leaning against a wall is 60� and the foot of the ladder is 4.6 m away from the wall. The length of the ladder is:

A.

B.

4.6 m

C.

7.8 m

D.

9.2 m

Answer with explanation

Answer: Option DExplanation

cos 60 = 4.6/X where X is the length of the ladder

X = 4.6/cos 60

= 4.6 divided by 1/2

= 9.2

I checked the answer as follows:

By pythagorean theorem, the ladder is 7.967433… feet off the ground.

Sin 60 = 7.967433/9.2

9.2

Workspace

A.

4√3 units

B.

8 units

C.

12 units

D.

Data inadequate

Answer with explanation

Answer: Option DExplanation

One of AB, AD and CD must have given.

So, the data is inadequate.

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

A.

173 m

B.

200 m

C.

273 m

D.

300 m

Answer with explanation

Answer: Option CExplanation

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.

0.63 meter/sec

B.

2.16 meter/sec

C.

3.87 meter/sec

D.

0.72 meter/sec

Answer with explanation

Answer: Option AExplanation

Workspace

The Top of a 15 metre high tower makes an angle of elevation of 60 degree with the bottom of an electric pole and angle of elevation of 30 degree with the top of pole. Find the height of the electric pole.

A.

7 metre

B.

8 metre

C.

9 metre

D.

10 metre

Answer with explanation

Answer: Option DExplanation

Workspace

A.

22 Km/Hr

B.

28 Km/Hr

C.

32 Km/Hr

D.

36 Km/Hr

Answer with explanation

Answer: Option CExplanation

Workspace

From a point C on a level ground, the angle of elevation of the top of a tower is 30 degree. If the tower is 100 meter high, find the distance from point C to the foot of the tower.

A.

170 meter

B.

172 meter

C.

173 meter

D.

167 meter

Answer with explanation

Answer: Option CExplanation

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

A.

276 meter

B.

273 meter

C.

270 meter

D.

263 metre

Answer with explanation

Answer: Option BExplanation

Workspace

A.

40sq.ft.

B.

800sq.ft.

C.

1600sq.ft.

D.

40/ sqrt 2 sq.ft.

Answer with explanation

Answer: Option BExplanation

Tan 60° = 3 = | PQ |

QR |

∴ PQ = 3 QR

Tan 30° = | 1 | = | PQ | = | PQ |

3 | SQ | 80+QR |

∴ 80 + QR = 3 PQ

∴ 80 + QR = 3QR

∴ QR = 40 ft.

If we read carefully, we see that the boy (Point R) and the scarecrow (Point Q) are in diagonally opposite corners.

So QR is a diagonal of the square farm.

Diagonal of square = side x 2

∴ 40 = side x 2

∴ side = 40/2

**∴ Area =** (side)^{2} = (40/2)^{2} = (1600/2) **= 800sq.ft.**

Workspace

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

Workspace

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

Workspace

^{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

Workspace

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

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.

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

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