**Word Problems Using Law of Sines and Cosines :**

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

**Question 1 :**

A researcher wants to determine the width of a pond from east to west, which cannot be done by actual measurement. From a point P, he finds the distance to the eastern-most point of the pond to be 8 km, while the distance to the western most point from P to be 6 km. If the angle between the two lines of sight is 60°, find the width of the pond.

**Solution :**

By finding the missing side, we may find the width of pond.

AP = a, PB = b and AB = c

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

cos 60 = (6^{2} + 8^{2} - c^{2}) / 2(6)(8)

1/2 = (36 + 64- c^{2}) / 96

48 = 100 - c^{2}

c^{2} = 100 - 48

c^{2} = 52

c = √52

= 2 √13 km

Hence the width of pond is 2 √13 km.

**Question 2 :**

Two Navy helicopters A and B are flying over the Bay of Bengal at same altitude from the sea level to search a missing boat. Pilots of both the helicopters sight the boat at the same time while they are apart 10 km from each other. If the distance of the boat from A is 6 km and if the line segment AB subtends 60° at the boat, find the distance of the boat from B.

**Solution :**

To find the missing side, we have to use the cosine formula.

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

Here we use the formula for cos C, because we know the length of b and c.

cos 60 = (a^{2} + 6^{2} - 10^{2})/2(6)(10)

1/2 = (a^{2} -64)/2a(6)

1 = a^{2} - 64/6a

a^{2} - 64 = 6a

a^{2} - 6a - 64 = 0

= (-b ± √b^{2} - 4ac)/2a

= (6 ± √(36 - 4(1)(-64))/2(1)

= (6 ± √292)/2

= (6 ± 2√73)/2

= 3 ± √73

Hence the distance from the helicopter B to boat is 3 + √73 km.

**Question 3 :**

A straight tunnel is to be made through a mountain. A surveyor observes the two extremities A and B of the tunnel to be built from a point P in front of the mountain. If AP = 3km, BP = 5 km and ∠APB = 120^{◦}, then find the length of the tunnel to be built

**Solution :**

Given that :

AP = 3 km = b, BP = 5 km = a and ∠APB = 120^{◦}

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

cos 120 = (5^{2} + 3^{2} - c^{2})/2(5)(3)

cos 120 = cos (90+30) = sin 30 = -1/2

-1/2 = (25 + 9 - c^{2})/30

-15 = 34 - c^{2}

c^{2 } = 49, c = 7

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

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