LAW OF SINES AND COSINES WORD PROBLEMS

Problem 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

Problem 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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