continued part of Some Applications of Trigonometry -Class 10

**10.Two poles of equal heights are standing opposite each other and either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30º, respectively. Find the height of poles and the distance of the point from the poles.**

Solution:

Let AB and CD be the poles and O is the point from where the elevation angles are measured.

In ∆ABO,

^{AB}/_{BO} = tan 60˚

^{AB}/_{BO} = √3

BO = ^{AB}/_{√3}

In ∆CDO,

^{CD}/_{DO} = tan 60˚

^{CD}/_{80-BO} = ^{1}/_{√3 }

CD√3 = 80 – BO

CD√3 = 80 – ^{AB}/_{√3}

CD√3 + ^{AB}/_{√3 }= 80

Since the poles are of equal heights,

CD = AB

CD[√3 + ^{1}/_{√3}] = 80

CD[^{√3 + 1}/_{√3}] = 80

CD = 20 √3m

BO = ^{AB}/_{√3} = ^{20√3}/_{√3 }= 20 m

DO = BD − BO = (80 − 20) m = 60 m

Therefore, the height of poles is 20√3 m and the point is 20 m and 60 m far from these poles.

**A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30°. Find the height of the tower and the width of the canal.**

**Solution:**

In ∆ABC,

^{AB}/_{BC} = tan60˚

^{AB}/_{BC} = √3

BC = ^{AB}/_{√3}

In ∆ABD,

^{AB}/_{BD} = tan30˚

^{AB}/_{BC+CD} = ^{1}/_{√3}

^{AB}/[_{(AB/√3)+20]} = ^{AB}/_{√3}

^{AB√3}/[_{(AB+20√3]} = ^{1}/_{√3}

3AB = AB + 20√3

2AB = 20√3

AB = 10√3 m

BC = ^{AB}/_{√3} = (^{10}^{√3}/_{√3})m = 10m

Therefore, the height of the tower is 10√3 m and the width of the canal is 10 m.

- From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.

Solution:

Let AB be a building and CD be a cable tower.

In ∆ABD,

^{AB}/_{BD} = tan45˚

^{7}/_{BD} = 1

BD = 7M

In ∆ACE,

AC = BD = 7 m

^{CE}/_{AE} = tan60˚

^{CE}/_{7} = √3

CE = 7√3

CD = CE + ED = 7√3m + 7m = 7(√3+1)m

Therefore, the height of the cable tower is 7(√3 + 1) m.

**As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.**

Solution:

Let AB be the lighthouse and the two ships be at point C and D respectively. In ∆ABC,

,

^{AB}/_{BC} = tan45˚

^{75}/_{BC} = 1

BC = 75 m

In ∆ABD

^{AB}/_{BD} = tan30˚

^{75}/_{75+CD} = ^{1}/_{√3}

75√3= 75 + CD

75(√3 – 1)m = CD

Therefore, the distance between the two ships is 75(√3 − 1) m

**A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After some time, the angle of elevation reduces to 30°. Find the distance traveled by the balloon during the interval.**

Solution:

Let the initial position A of balloon change to B after some time and CD be the girl.

In ∆ACE,

^{AE}/_{CE} = tan60˚

^{AF-EF}/_{CE} = √3

^{88.2-1.2}/_{CE} = √3

CE = ^{87}/_{√3} = 29√3m

In ∆BCG,

^{BG}/_{CG} = tan30˚

^{88.2-1.2}/_{CG} = ^{1}/_{√3}

87√3m = CG

Distance travelled by balloon = EG = CG – CE

= (87√3 – 29√3)m

= 58√3m

- A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car as an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.

Solution:

Let AB be the tower. Initial position of the car is C, which changes to D after six seconds.

In ∆ADB,

^{AB}/_{DB} = tan60˚

^{AB}/_{DB} = √3

DB = ^{AB}/_{√3}

In ∆ABC,

^{AB}/_{BC} = tan30˚

^{AB}/_{BC+CD} = ^{1}/_{√3}

AB√3= BD + DC

AB√3=^{ AB/}_{√3} + DC

DC = AB√3 – ^{AB}/_{√3} = AB(√3-^{1}/_{√3})

= ^{2AB}/_{√3}

Time taken by the car to travel distance DC = (= ^{2AB}/_{√3}) 6 seconds,

time taken by the car to travel distance DB (= ^{2AB}/_{√3}) = ^{6}/_{(2AB/√3)} X ^{AB}/_{√3} = ^{6}/_{2} = 3 seconds

**The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m. from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.**

Solution:

Let AQ be the tower and R, S are the points 4m, 9m away from the base of the tower respectively. The angles are complementary. Therefore, if one angle is θ, the other will be 90 − θ.

In ∆AQR,

^{AQ}/_{QR} = tan θ

^{AQ}/_{4} = tanθ ———–(i)

In ∆AQS,

^{AQ}/_{SQ} = tan(90- θ)

^{AQ}/_{9} = cot θ ———–(II)

On multiplying equations (i) and (ii), we obtain

^{AQ}/_{4 }. ^{AQ}/_{9} = cot θ. tanθ

AQ^{2}/_{36} = 1

AQ^{2} = 36

AQ = √36 = ±6

However, height cannot be negative.

Therefore, the height of the tower is 6 m.

I love Trigonometry 🙂 This post reminds me of my childhood

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