LAWS OF SINES AND COSINES PRACTICAL PROBLEMS

Problem 1 :

A plane is 1 km from one landmark and 2 km from another. From the planes point of view the land between them subtends an angle of 45°. How far apart are the landmarks?

Solution :

We have to find the length of AB.

a = 2 km (Opposite to angle A)

b = 1 km (Opposite to angle B)

c = AB (Opposite to angle C)

Cosine formula :

cos C  =  (a2 + b2 - c2) / 2ab

cos 45  =  (22 + 12 - c2) / 2(2)(1)

1/√2  =  (5 - c2) / 4

- c=  4/√2

5 - c2  =  2√2

c2  =  2√2 + 5

c  =  √(2√2 + 5)

Problem 2 :

A man starts his morning walk at a point A reaches two points B and C and finally back to A such that ∠A = 60 and ∠B = 45, AC = 4 km in the triangle ABC. Find the total distance he covered during his morning walk.

Solution :

<A + <B + <C  =  180

60 + 45 + <C  =  180

<C  =  180 - 105  =  75

a/sin A  =  b/sin B  =  c/sin C

a/sin 60  =  4/sin 45  =  c/sin 75  ----(1)

First let us find "a".

a/sin 60  =  4/sin 45

a/(√3/2) =  4/(1/√2)

2a/√3  =  4√2

a  =  4√2√3/2

a  =  2√6

By applying a  =  2√6 in (1) and find c.

2√6/sin 60  =  c/sin 75 

2√6/(√3/2)  =  c/sin 75 

sin 75  =  sin (45 + 30)

  =  sin 45 cos 30 + cos 45 sin 30

  =  (1/√2) (√3/2) + (1/√2)(1/2)

  =  (√3 + 1)/2√2

4√2  =  c/(√3 + 1)/2√2

4√2  =  2√2c/(√3 + 1)

c  =  4√2(√3 + 1)/ 2√2

c  =  2(√3 + 1)

Total distance covered  =  2√6 + 4 + 2(√3 + 1)

 =  2√6 + 4 + 2√3 + 2

 =  2√6 + 6 + 2√3

Problem 3 :

Two vehicles leave the same place P at the same time moving along two different roads. One vehicle moves at an average speed of 60 km/hr and the other vehicle moves at an average speed of 80 km/hr. After half an hour the vehicle reach the destinations A and B. If AB subtends 60◦ at the initial point P, then find AB. 

Solution :

Distance covered by A  =  time x speed 

  =  (1/2) ⋅ 60 

  =  30 km

Distance covered by B  =  time x speed 

  =  (1/2) ⋅ 80 

  =  40 km

cos C  =  a2 + b2 - c2/2ab

cos 60 =  402 + 302 - c2/2(40)(30)

1/2  =  (1600 + 900 - c2)/2400

1200  =  2500 - c2

c =  2500 - 1200  =  1300

c = √1300

c  =  10√13

Hence the distance of A and B is  10√13 km.

Problem 4 :

Suppose that a satellite in space, an earth station and the center of earth all lie in the same plane. Let r be the radius of earth and R be the distance from the center of earth to the satellite. Let d be the distance from the earth station to the satellite. Let 30 be the angle of elevation from the earth station to the satellite. If the line segment connecting earth station and satellite subtends angle α at the center of earth , then prove that d = R1 + (r/R)− 2 (r/R) cos α.

Solution :

Let A be the center, B be the station and C be the satellite position.

a/sin A  =  b/sin B  =  c/sin C

a = d, b = R and c = r

d/sin a  =  R/sin 30  = r/sin (180 - (30 +a))

cos A  =  (b2 + c2 - a2)/2bc

cos a  =  (R2 + r2 - d2)/2Rr

2Rr cos a  =  R2 + r2 - d2

d2  =  R2 + r- 2Rr cos a

d  =  √(R2 + r- 2Rr cos a)

d  =  √(R2 (1 + (r/R)- 2(r/R) cos a

d  =  R(1 + (r/R)- 2(r/R) cos a)

Hence proved.

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