Test: Trigonometry- 2 - GMAT MCQ

Let BD be the lighthouse and A and C be the positions of the ships.
Then, BD = 100 m,  BAD = 30° ,  BCD = 45°


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

Two equations and 3 variables. Hence we can not find the required value with the given data.

(Note that if one of SR, PQ, QR is known, this becomes two equations and two variables and if that was the case, we could have found out the required value.)

PQ=200√3=200×1.73=346 m

Consider the diagram shown above where QR represents the tree and PQ represents its shadow
We have, QR = PQ
Let QPR = θ

i.e., required angle of elevation = 45°

SR = PQ = 2 m
PS = QR =10√3m
 
TR = TS + SR = 10 + 2 = 12 m

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

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

Then, ∠ADC = 30° , ∠ACB = 45°

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



⇒ y = √3h−h  (∵ Substituted the value of x from equation 1 )
⇒y = h(√3−1)
Given that distance y is covered in 8 minutes.
i.e, distance h(√3−1) is covered in 8 minutes.
Time to travel distance x
= Time to travel distance h (∵ 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

Consider the diagram shown above. 
Let AB be the tower. Let C and D be the positions of the boat
Then, ACB = 45° , ADC = 30°, BC = 100 m


(∵ Substituted the value of AB from equation 1)

=> BD =100√3

CD = (BD - BC) =(100√3−100)=100(√3−1)
It is given that the distance CD is covered in 10 seconds.
i.e., the distance 100(√3−1) is covered in 10 seconds.

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


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

i.e., height of the electric pole = 10 m

Let DC be the tower and A and B be the positions of the observer such that AB = 40 m
We have DAC = 30°, DBC = 45°
Let DC = h


We know that, AB = (AC - BC)
=> 40 = (AC - BC)
=> 40=(h√3−h)40= [∵ from (1) & (2)]
=>40=h(√3−1)


=20(1.73+1)=20×2.73=54.6 m

Let DC be the tower and A and B be the objects as shown above.
Given that DC = 600 m , DAC = 45°, DBC = 60°


Distance between the objects 
= AB = (AC - BC)

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 , BAC = 60°

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, DAB = 30° 
ABC = DAB = 30°   (because DA || BC)


=80×1.73=138.4 mtan⁡30°=ACBC=>tan⁡30°=80BC=>BC = 80tan⁡30°=80(13)=80×1.73=138.4 m
i.e., Distance of the bus from the foot of the tower = 138.4 m

Let BD be the lighthouse and A and C be the two points on ground.
Then, BD, the height of the lighthouse = 60 m
 BAD = 45° ,  BCD = 60°



Distance between the two points A and C 
= AC = BA + BC 
= 60 + 34.6   [∵ Substituted value of BA and BC from (1) and (2)]
= 94.6 m

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

Given that AC = 100 m 
∠XAD = ∠ADB = 30°   (∵ AX || BD )
∠XAE = ∠AEC = 60°   (∵ AX || CE)
Let DE = h
Then, BC = DE = h, 
AB = (100-h)   (∵ AC=100 and BC = h), 
BD = CE

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

⇒ h=100−33.33=66.67 m
i.e., the height of the pole = 66.67 m