LAW OF SINES AND COSINES WORD PROBLEMS

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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)

Problem 3 :

The course of a boat race starts at point A in the figure and proceeds in the direction S 52^◦W to point B, then the directions S 40^◦E to point C, and finally back to A point C lies 8 kilometers directly south of point A. Approximate the total distance of the race course.

Solution :

Lines AC and BD are parallel, it follows <BCA = <CBD consequently triangle ABC has the measures in the figure.

The measure of the angle B is

= 180 - (52 + 40)

= 180 - 92

= 88

a/sin 52 = b/sin 88 = c/sin 40

Applying b = 8

a/sin 52 = 8/sin 88 = c/sin 40

a/sin 52 = 8/sin 88

a = 8 sin 52 / sin 88

= 8(0.788)/0.999

= 6.304/0.999

= 6.31

8/sin 88 = c/sin 40

c = 8(sin 40)/sin 88

= 8(0.642)/0.999

= 5.136/0.999

= 5.14

The total length of the course

= 8 + 6.31 + 5.14

= 19.45 km (approximately)

Problem 4 :

Because of prevailing winds, a tree grew so that it was leaning 4 degree from the vertical. At the point 40 meters from the tree, the angle of elevation to the top of the tree is 30 degree. Find the height of the tree.

Solution :

<B = 94, <C = 30, <A = 180 - (94 + 30)

= 180 - 124

= 56

BC/sin 56 = AC/sin 94 = AB/sin30

40/sin 56 = AC/sin 94 = h/sin30

40/sin 56 = h/sin30

h = 40(sin 30)/sin 56

= 40(0.5)/0.829

= 20/0.829

= 24.12

So, the required distance is 24.12 meters.

Problem 5 :

A plane flies 500 kilometers with bearing of 316° from Naples to Elgin. The plane then flies 720 kilometers from Elgin to Canton (Canton is due west to Naples). Find the bearing of the flight from Elgin to Canton.

Solution :

<C = 90 - 44

= 46

In triangle ABC, using sin law

a/sin A = b/sin B = c/sin C

500/sin B = 720/sin 46

sin B = 500(sin 46)/720

= 500(0.719)/720

sin B = 0.499

B = sin^-1(0.499)

B = 29.93

Let y be the small part.

<A = 180 - (29.93 + 46 + 44)

= 180 - 119.93

= 60.03

Problem 6 :

A bridge is to be built across a small lake from gazebo to a dock shown in the figure. The bearing from gazebo to the dock is S41° W. Form a tree 100 meters from gazebo, the bearings to the gazebo and dock are S74°E and S28°E respectively. Find the distance from gazebo to the dock.