Oblique Triangles: Law of Sines

OBLIQUE TRIANGLES: (no right angle)

The techniques for solving triangles we discussed in lesson 1 apply only to right angled triangles We use two laws -- the Law of Sines, and the Law of Cosines

to solve oblique or non-right triangles.

In order to use the Law of Sines, the given information must include the measure of an angle and the measure of the side opposite that angle. If we have two side lengths and the measure of the contained angle, or just the 3 side lengths, we must use the Law of Cosines.

Remember that these laws apply to oblique triangles. If you have a right triangle, use the techniques derived from the trig function definitions in lesson 1.

THE LAW OF SINES:

The Law of Sines includes two different statements used to solve for the measures of the sides or the angles in an oblique triangle. Since it is best to place the unknown in the numerator of a fraction:

If you are solving for an unknown side, use: where a is the side opposite angle A etc.

If you are solving for an unknown angle, use the reciprocal statements: Example 1: We're given 2 angles and 1 side.

Solve the triangle completely: let's find angle C: 180° ( 30° + 45° ) = 105°

now, , so b = = 5.66 c = 7.73.

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Now let's do one where we have to find an unknown angle.

Example 2: We're given 2 sides and 1 angle. Since we know a, we can solve for angle A so sin A = = 0.7075

To find angle A we take arcsin (0.7075) = 45.03 ° .

angle C = 180° (30° + 45.03°) = 104.96°

so, to find c, c = 7.73

The Ambiguous Case

When we use the Law of Sines, if we are given the measure of one angle, the length of the side opposite that angle and another side, there could be 2 triangles to solve.

Here's an example.

Example 3: the ambiguous case

Solve triangle ABC where B = 37.2º , b = 16.4 cm and a = 22.3 cm.

We construct the triangle like this:

First, make angle B at 37.2º. By this, we have drawn segments of sides a and c.

We now measure 22.3 cm along side a to define vertex C.

Now, we set our compasses to 16.4 cm, place the point at C and draw an arc which cuts the segment of side c in 2 places. Both these points are vertex A.

Generally, we label one of them A and the other A 1 . Side A 1B will be labeled c1 and side AB will be labeled c.

Why does this happen?

The common sense answer is that there are 2 places where an arc with radius 16.4 cm cuts the line AB. The mathematical answer is that the sine of an angle is positive in both the 1st and 2nd quadrants

so, angle A1 is 180º – angle A, the acute angle our calculators display when we solve
for the measure of angle A.

Here's the solution: angle BAC = arcsin (0.82211) = 55.3º

This means that angle BA1C = 180º – 55.3º = 124.7º

The best thing to do at this point is to redraw the diagram with the two triangles shown separately. Another approach is to use colors as in the diagram.

We see a small triangle A 1BC outlined in green, and triangle ABC outlined in pink.

Now that we know 2 angles and 2 sides in both triangles, we solve for the missing info using the Law of Sines as we did in the previous examples.

Note: If angle B is obtuse ( > 90º ) there is only one triangle to solve.

Now get a pencil, an eraser and a note book, copy the questions,
do the practice exercise(s), then check your work with the solutions.
If you get stuck, review the examples in the lesson, then try again. Practice

Make a diagram for questions where one is not provided.

1) Use the Law of Sines to solve the triangles ABC:

a) b = 15.43, a = 21.76, angle B = 28.25° ( solution )

b) angle B = 37.2°, b = 16.4 cm a = 22.3 cm ( solution )

c) a = 4.25 m b = 7.58 m angle A = 22.67° ( solution )

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2) A building which is 16.07 m high stands at the top of a hill. From the top of the building, the angle of depression to a car at the bottom of the hill is 55.3°. From the bottom of the building, the angle of depression to the same car is 12.38°. How far is the car from the bottom of the building? ( solution )

3) Two observers who are 50 meters apart see an object falling between them and the reported angles of elevation are 32.37° and 43.2°. How high is the falling object above the ground?

( solution )

4) Points A and B are on opposite sides of a lunar crater. Point C is 50 m from point A.
Angle BAC is 112° and angle ACB is 42°. How wide is the lunar crater? ( solution )

5) A wire supporting a pole makes a 71° angle with the level ground. At a point 7.62 m away from the foot of the wire, the angle of elevation to the top of the pole is 37° . How long is the wire? ( solution )

6) A vertical pole stands by a road that is inclined 10° to the horizontal. When the angle of elevation of the sun is 23°, the pole's shadow is 11.67 m long directly downhill along the road. How tall is the pole? ( solution )

. Solutions

1) Use the Law of Sines to solve the triangles ABC:

a) b = 15.43, a = 21.76, angle B = 28.25°  angle A = arcsin 0.667494 angle A = 41.87°

angle C = 180° – ( 41.87° + 28.25°) = 109.88° = 30.66

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b) angle B = 37.2°, b = 16.4 cm a = 22.3 cm  angle A = arcsin 0.822107 angle A = 55.3°

angle C = 180° – (55.3° + 37.2°) = 87.5° = 27.1 cm.

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c) a = 4.25 m b = 7.58 m angle A = 22.67°  angle B = arcsin 0.687413 = 43.43°

angle C = 180° – (43.43° + 22.67°) = 113.9° c = 10.1 m.

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2) A building which is 16.07 m high stands at the top of a hill. From the top of the building, the angle of depression to a car at the bottom of the hill is 55.3°. From the bottom of the building, the angle of depression to the same car is 12.38°. How far is the car from the foot of the building? angle DAC = 90°– 55.3° = 34.7° b = 13.43 m.

The car is 13.43 m from the foot of the building.

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3) Two observers who are 50 meters apart see an object falling between them and the reported angles of elevation are 32.37° and 43.2°. How high is the falling object above the ground? angle AOB = 180° (32.37° + 43.2°) = 104.43° b = 35.34 m

sin 32.37° = h = 35.34 (sin 32.37°) = 18.92 m

The object is 18.92 m high.

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4) Points A and B are on opposite sides of a lunar crater. Point C is 50 m from point A.
Angle BAC is 112° and angle ACB is 42°. How wide is the lunar crater? angle B = 180° – (112° + 42°) = 26° c = 76.32 m

The lunar crater is 76.32 m wide.

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5) A wire supporting a pole makes a 71° angle with the level ground. At a point 7.62 m away from the foot of the wire, the angle of elevation to the top of the pole is 37°.
How long is the wire? angle PWA = 180° – 71° = 109°

angle WPA = 180° – (109° + 37°) = 34° x = 8.2 m

The wire is 8.2 m long.

6) A vertical pole stands by a road that is inclined 100 to the horizontal. When the angle of elevation of the sun is 230, the pole's shadow is 11.67 m long directly downhill along the road. How tall is the pole? angle A = 180° – (100° + 23°) = 57° c = 5.44 m

The pole is 5.44 m tall.

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