**Laws of Sine and Cosine Practical Problems :**

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

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

cos 45 = (2^{2} + 1^{2} - c^{2}) / 2(2)(1)

1/√2 = (5 - c^{2}) / 4

5 - c^{2 }= 4/√2

5 - c^{2 }= 2√2

c^{2} = 2√2 + 5

c = √(2√2 + 5)

**Question 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

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

cos 60 = 40^{2} + 30^{2} - c^{2}/2(40)(30)

1/2 = (1600 + 900 - c^{2})/2400

1200 = 2500 - c^{2}

c^{2 } = 2500 - 1200 = 1300

c = √1300

c = 10√13

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

**Question 4 :**

Suppose that a satellite in space, an earth station and the centre of earth all lie in the same plane.Let r be the radius of earth and R be the distance from the centre 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 centre of earth , then prove that d = R√1 + (r/R)^{2 }− 2 (r/R) cos α.

Let A be the centre, 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 = (b^{2} + c^{2} - a^{2})/2bc

cos a = (R^{2} + r^{2} - d^{2})/2Rr

2Rr cos a = R^{2} + r^{2} - d^{2}

d^{2} = R^{2} + r^{2 }- 2Rr cos a

d = √(R^{2} + r^{2 }- 2Rr cos a)

d = √(R^{2} (1 + (r/R)^{2 }- 2(r/R) cos a

d = R√(1 + (r/R)^{2 }- 2(r/R) cos a)

Hence proved.

After having gone through the stuff given above, we hope that the students would have understood, "Laws of Sine and Cosine Practical Problems"

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