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**1. 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 30Â° and 45Â° 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: C**Explanation:**

Let AB be the lighthouse and C and D be the positions of the ships.

**2. A man standing at a point P is watching the top of a tower, which makes an angle of elevation of 30Âº with the man's eye. The man walks some distance towards the tower to watch its top and the angle of the elevation becomes 60Âº. What is the distance between the base of the tower and the point P?**

A. 43 units

B. 8 units

C. 12 units

D. Data inadequate

E. None of these**Answer: DExplanation:**

One of AB, AD and CD must have given.

So, the data is inadequate.**3. The angle of elevation of a ladder leaning against a wall is 60Âº and the foot of the ladder is 4.6 m away from the wall. The length of the ladder is:**

A. 2.3 m

B. 4.6 m

C. 7.8 m

D. 9.2 m**Answer: D****Explanation:**

Let AB be the wall and BC be the ladder.

**4. An observer 1.6 m tall is 20âˆš3 away from a tower. The angle of elevation from his eye to the top of the tower is 30Âº. The heights of the tower is:**

A.21.6 m

B. 23.2 m

C. 24.72 m

D. None of these**Answer: AExplanation:**

Let AB be the observer and CD be the tower.

**5. From a point P on a level ground, the angle of elevation of the top tower is 30Âº. 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: C****Explanation:**

Let AB be the tower.

**6. The angle of elevation of a ladder leaning against a wall is 60Âº and the foot of the ladder is 12.4 m away from the wall. The length of the ladder is:**

A. 14.8 m

B. 6.2 m

C. 12.4 m

D. 24.8 m**Answer: D****Explanation:**

Consider the diagram shown above where PR represents the ladder and RQ represents the wall.

cos 60Â° = PQ/PR

1/2 = 12.4/PR

PR = 2 Ã— 12.4 = 24.8 m**7. A man on the top of a vertical observation tower observers a car moving at a uniform speed coming directly towards it. If it takes 8 minutes for the angle of depression to change from 30Â° to 45Â°, how soon after this will the car reach the observation tower?**

A. 8 min 17 second

B. 10 min 57 second

C. 14 min 34 second

D. 12 min 23 second**Answer: B****Explanation:**

Consider the diagram shown above. Let AB be the tower. Let D and C be the positions of the car.

Then, âˆ ADC = 30Â° , âˆ ACB = 45Â°

Let AB = h, BC = x, CD = y

tan45Âº = AB/BC = h/x

â‡’ 1 = h/x

â‡’ h = x .... (1)

tan 30Âº = AB/BD = AB/(BC + CD) = H/x+y

â‡’ 1âˆš3 = h/x + y

â‡’ x + y = âˆš3h

â‡’ y = âˆš3h - x

â‡’ y = âˆš3h - h (âˆµ Substituted the value of x from equation 1 )

â‡’ y = h(âˆš3 - 1)

Given that distance y is covered in 8 minutes.

i.e, distance **h(âˆš3 - 1)** is covered in 8 minutes.

Time to travel distance x

= Time to travel distance h (âˆµ Since x = h as per equation 1).

Let distance h is covered in t minutes.

since distance is proportional to the time when the speed is constant, we have

â‰ˆ 10 minutes 57 seconds**8. A man is watching from the top of a tower a boat speeding away from the tower. The boat makes an angle of depression of 45Â° with the man's eye when at a distance of 100 metres from the tower. After 10 seconds, the angle of depression becomes 30Â°. What is the approximate speed of the boat, assuming that it is running in still water?**

A. 26.28 km/hr

B. 32.42 km/hr

C. 24.22 km/hr

D. 31.25 km/hr**Answer: A****Explanation:**

Consider the diagram shown above.

Let AB be the tower. Let C and D be the positions of the boat

Then, angleACB = 45Â° , angleADC = 30Â°, BC = 100 m

(âˆµ Substituted the value of AB from equation 1)

CD = (BD - BC) = (100 âˆš3 âˆ’ 100) = 100 (âˆš3 âˆ’ 1)

It is given that the distance CD is covered in 10 seconds. i.e., the distance 100(âˆš3 âˆ’ 1) is covered in 10 seconds.

Required speed

=Distance/Time

= 7.3 meter/seconds

= 7.3 Ã— 18/5 km/hr = 26.28 km/hr.**9. The top of a 15 metre high tower makes an angle of elevation of 60Â° with the bottom of an electronic pole and angle of elevation of 30Â° with the top of the pole. What is the height of the electric pole?**

A. 5 metres

B. 8 metres

C. 10 metres

D. 12 metres**Answer: C****Explanation:**

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

Given that AC = 15 m, angleADB = 30Â°, angleAEC = 60Â°

Let DE = h

Then, BC = DE = h,

AB = (15-h) (âˆµ AC=15 and BC = h),

BD = CE

(âˆµ BD = CE and substituted the value of CE from equation 1)

=> h = 15 âˆ’ 5 = 10 m

i.e., height of the electric pole = 10 m**10. The angle of elevation of the top of a tower from a certain point is 30Â°. If the observer moves 40 m towards the tower, the angle of elevation of the top of the tower increases by 15Â°. The height of the tower is:**

A. 64.2 m

B. 62.2 m

C. 52.2 m

D. 54.6 m**Answer: D****Explanation:**

Let DC be the tower and A and B be the positions of the observer such that AB = 40 m

We have âˆ DAC = 30Â°, âˆ DBC = 45Â°

Let DC = h

tan 30Â° = DC/AC

tan 45Â° = DC/BC

=> 1 = h/BC

=> BC = h â‹¯(2)

We know that, AB = (AC - BC)

=> 40 = (AC - BC)

=> 40 = ( hâˆš3 âˆ’ h) [âˆµ from (1) & (2)]

=> 40 = h (âˆš3 âˆ’ 1)

= 20 (âˆš3 + 1) = 20 (1.73 + 1) = 20 Ã— 2.73 = 54.6 m**11. On the same side of a tower, two objects are located. Observed from the top of the tower, their angles of depression are 45Â° and 60Â°. If the height of the tower is 600 m, the distance between the objects is approximately equal to :**

A. 272 m

B. 284 m

C. 288 m

D. 254 m**Answer: D****Explanation: **

Let DC be the tower and A and B be the objects as shown above.

Given that DC = 600 m , âˆ DAC = 45Â°, âˆ DBC = 60Â°

tan 60Â° = DC/BC

âˆš3 = 600/BC

BC = 600âˆš3 â‹¯ ( 1 )

tan 45Â° = DC/AC

1 = 600/AC

AC= 600 â‹¯ ( 2 )

Distance between the objects

= AB = (AC - BC)

= 600 âˆ’ 600/âˆš3 [âˆµ from (1) and (2)]

= 200 âˆš3(âˆš3 âˆ’ 1)

= 200(3 âˆ’ âˆš3)

= 200 (3 âˆ’ 1.73)

= 254 m**12. A ladder 10 m long just reaches the top of a wall and makes an angle of 60Â° with the wall.Find the distance of the foot of the ladder from the wall (âˆš3 = 1.73)**

A. 4.32 m

B. 17.3 m

C. 5 m

D. 8.65 m**Answer: DExplanation:**

Let BA be the ladder and AC be the wall as shown above. Then the distance of the foot of the ladder from the wall = BC

Given that BA = 10 m , âˆ BAC = 60Â°

sin 60Â° = BC/BA

âˆš3/2 = BC/10

BC = 10 Ã— âˆš3/2 = 5 Ã— 1.73 = 8.65 m

A. 40 m

B. 138.4 m

C. 46.24 m

D. 160 m

Explanation:

Let AC be the tower and B be the position of the bus. Then BC = the distance of the bus from the foot of the tower.

Given that height of the tower, AC = 80 m and the angle of depression, âˆ DAB = 30Â°

âˆ ABC = âˆ DAB = 30Â° (because DA || BC)

tan 30Â° = AC/BC

=> tan 30Â° = 80/BC

= 80 Ã— 1.73 = 138.4 m

i.e., Distance of the bus from the foot of the tower = 138.4 m

A. 45 m

B. 30 m

C. 103.8 m

D. 94.6 m

Explanation:

Let BD be the lighthouse and A and C be the two points on ground. Then, BD, the height of the lighthouse = 60 m

âˆ BAD = 45Â° , âˆ BCD = 60Â°

tan 45Â° = BD/BA

â‡’ 1 = 60/BA

â‡’ BA = 60 m â‹¯ ( 1 )

tan 60Â° = BD/BC

â‡’ âˆš3 = 60/BC

Distance between the two points A and C

= AC = BA + BC

= 60 + 34.6 [âˆµ Substituted value of BA and BC from (1) and (2)]

= 94.6 m

A. 52 m

B. 50 m

C. 66.67 m

D. 33.33 m

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

Given that AC = 100 m

âˆ XAD = âˆ ADB = 30Â° (âˆµ AX || BD)

âˆ XAE = âˆ AEC = 60Â° (âˆµ AX || CE)

Let DE = h

Then, BC = DE = h,

AB = (100-h) (âˆµ AC=100 and BC = h),

BD = CE

(âˆµ BD = CE and substituted the value of CE from equation 1)

= 100/3 = 33.33

=> h = 100 âˆ’ 33.33 = 66.67 m

i.e., the height of the pole = 66.67 m

A. 9 m

B. 10.40 m

C. 15.57 m

D. 12 m

Let DC be the vertical tower and AD be the vertical flagpole. Let B be the point of observation.

Given that AD = 18 m, âˆ ABC = 60Â°, âˆ DBC = 30Â°

Let DC be h.

tan 30Â° = DC/BC

1âˆš3 = h/BC

h = BC/âˆš3â‹¯ (1)

tan 60Â° = AC/BC

18 + h = BC Ã— âˆš3 â‹¯(2)

=> 3 h = 18 + h

=> 2 h = 18

=> h = 9 m

i.e., the height of the tower = 9 m**17. 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 60Â°. 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: B****Explanation:****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, âˆ BCA = 60Â° **

tan60Â°= BA/CA

âˆš3 = BA/150

BA = 150âˆš3

i.e, the distance travelled by the balloon = 150âˆš3 meters time taken = 2 min = 2 Ã— 60 = 120 seconds

Speed = Distance/Time

= 150âˆš3/120 = 1.25 âˆš3

= 1.25 Ã— 1.73 = 2.16 meter/second**18. The angles of depression and elevation of the top of a wall 11 m high from top and bottom of a tree are 60Â° and 30Â° respectively. What is the height of the tree?**

A. 22 m

B. 44 m

C. 33 m

D. None of these**Answer: B****Explanation:**

Let DC be the wall, AB be the tree.

Given that âˆ DBC = 30Â°, âˆ DAE = 60Â°, DC = 11 m

tan 30Â° = DC/BC

1âˆš3 = 11/BC

BC = 11âˆš3m

AE = BC = 11âˆš3 m â‹¯ (1)

tan 60 Â° = ED/AE

âˆš3 = ED/11âˆš3 [âˆµ Substituted value of AE from (1)]

ED = 11âˆš3 Ã— âˆš3 = 11 Ã— 3 = 33

Height of the tree

= AB = EC = (ED + DC)

= (33 + 11) = 44 m**19. Two vertical poles are 200 m apart and the height of one is double that of the other. From the middle point of the line joining their feet, an observer finds the angular elevations of their tops to be complementary. Find the heights of the poles.**

A. 141 m and 282 m

B. 70.5 m and 141 m

C. 65 m and 130 m

D. 130 m and 260 m**Answer: B****Explanation:**

Let AB and CD be the poles with heights h and 2h respectively.

Given that distance between the poles, BD = 200 m

Let E be the middle point of BD,

âˆ AEB = Î¸ âˆ CED = (90-Î¸) (âˆµ given that angular elevations are complementary)

Since E is the middle point of BD, we have BE = ED = 100 m

From the right Î”ABE,

tanÎ¸ = AB/BE

tanÎ¸ = h/100

h = 100 tan Î¸ â‹¯ ( 1 )

From the right Î”EDC,

=> âˆš2h = 100

= 50âˆš2 = 50 Ã— 1.41 = 70.5

=> 2h = 2 Ã— 70.5 = 141

i.e., the height of the poles are 70.5 m and 141 m.**20. To a man standing outside his house, the angles of elevation of the top and bottom of a window are 60Â° and 45Â° respectively. If the height of the man is 180 cm and he is 5 m away from the wall, what is the length of the window?**

A. 8.65 m

B. 2 m

C. 2.5 m

D. 3.65 m**Answer: D****Explanation:**

Let AB be the man and CD be the window

Given that the height of the man, AB = 180 cm, the distance between the man and the wall, BE = 5 m,

âˆ DAF = 45Â° , âˆ CAF = 60Â°

From the diagram, AF = BE = 5 m

From the right Î”AFD,

tan 45 Â° = DF/AF

1 = DF/5

DF = 5 â‹¯ ( 1 )

From the right Î”AFC,

tan 60 Â° = CF/AF

âˆš3 = CF/5

CF = 5âˆš3 â‹¯ ( 2 )

Length of the window

= CD = (CF - DF)

= 5 âˆš 3 âˆ’ 5 [âˆµ Substituted the value of CF and DF from (1) and (2)]

= 5 (âˆš3 âˆ’ 1) = 5 (1.73 âˆ’ 1)

= 5 Ã— 0.73 = 3.65 m

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