**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.

**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 = (a^{2} + b^{2} - c^{2}) / 2ab

cos 60^{◦} = (120^{2} + 60^{2} - c^{2}) / 2(120)(60)

1/2 = (14400 + 3600 - c^{2}) / 14400

7200 = 18000 - c^{2}

c^{2} = 18000-7200

c^{2} = 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)

After having gone through the stuff given above, we hope that the students would have understood, "Law of Sines and Cosines Word Problems"

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