# LAW OF SINES AND COSINES WORD PROBLEMS

## About "Law of Sines and Cosines Word Problems"

Law of Sines and Cosines Word Problems :

Here we are going to see some practical problems using the concept of law of sines and cosines.

## Word Problems Using Law of Sines and Cosines - Examples

Question 1 :

A farmer wants to purchase a triangular shaped land with sides 120 feet and 60 feet and the angle included between these two sides is 60. If the land costs Rs. 500 per sq.ft, find the amount he needed to purchase the land. Also find the perimeter of the land.

Solution :

To find the missing side, let us use cosine formula.

a = 120 feet, b = 60 feet

The angle included by the sides a and b will be C.

<C  =  60

cos C  =  (a2 + b2 - c2) / 2ab

cos 60  =  (1202 + 602 - c2) / 2(120)(60)

1/2  =  (14400 + 3600 - c2) / 14400

7200  =  18000 - c2

c2  =  18000-7200

c2  =  10800

c  =  60√3

Perimeter of the triangle = 120+60+60√3

=  180 + 60√3

Area of triangle  =  (1/2)ab sin C

=  (1/2) (120)(60) sin 60

=  60 (60)( 3/2)

=  3117.6

Cost of land  =  Rs 500 per square feet

Required cost  =  3117.6(500)

=  Rs. 1558800

Question 2 :

A fighter jet has to hit a small target by flying a horizontal distance. When the target is sighted, the pilot measures the angle of depression to be 30. If after 100 km, the target has an angle of depression of 45, how far is the target from the fighter jet at that instant?

Solution :

From the diagram, we come to know that we have to find the value of a.

Sine formula :

a/sin A  =  b/sin B  =  c/sin C

45 + <B  =  180

<B  =  180 - 45  =  135

<BCA  =  45 - 30  =  15

a/sin 30  =  b/sin 135  =  100/sin 15

a/sin 30  =   100/sin 15

sin (45 - 30)  =  sin 45 cos 30 - cos 45 sin 30

=  (1/√2) (√3/2) - (1/√2) (1/2)

=  (√3/2√2) - (1/2√2)

=  (√3 - 1)/2√2

a/(1/2)  =  100/((√3 - 1)/2√2)

2a  =  200√2/(√3 - 1)

a  =  [200√2/(√3 - 1)] [(√3 + 1)/(√3 + 1)]

=  200√2(√3 + 1)/2

a  =  50√2(√3 + 1)

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