11.06 Sine rule

Let's start by drawing a line segment from the vertex C perpendicular to the edge c. The length of this segment is labelled as x.

Since x is perpendicular to c, the two line segments meet at right angles. This means that the triangle above has been divided into two right-angled triangles.

Using basic trigonometry, the relationships for the sines of the angles A and B is given by \sin A = \dfrac{x}{b} and \sin B = \dfrac{x}{a}.

Multiply the first equation by b and the second by a. This rearranges the sine ratios above so that x is the subject of both equations, x = b \sin A and x = a \sin B.

Equating these two equations eliminates the x, creating the following relationship between angles A and B, and side-lengths a and b. b \sin A = a \sin B

Dividing this last equation by the side lengths a & b gives the following important relationship: \dfrac{\sin A}{a} = \dfrac{\sin B}{b}

This process can be repeated to find how these two angles relate to c and C, and this gives us the sine rule (sometimes called the law of sines).

For a triangle with sides a, b, and c, with corresponding angles A,\,B, and C: \dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}.

The reciprocal of each fraction gives the alternate form: \dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}.

The sine rule shows that the lengths of the sides in a triangle are proportional to the sines of the angles opposite them.

Suppose we have the angles A and B and a known side length b (opposite to angle B). The sine rule can be used to calculate unknown side length a (opposite to angle A). Using the form of the sine rule with numerator lengths \dfrac{a}{\sin A}=\dfrac{b}{\sin B}, this equation can be transposed to make a the subject of the equation, by multiplying both sides by \sin A. This gives a=\dfrac{b\sin A}{\sin B}.

Examples

Example 1

Find the side length a using the sine rule. Round your answer to two decimal places.

Worked Solution

Create a strategy

Use the sine rule: \dfrac{a}{\sin A}=\dfrac{b}{\sin B}.

Apply the idea

\displaystyle \frac{a}{\sin 33\degree}	\displaystyle =	\displaystyle \frac{18}{\sin 69\degree }	Substitute b=18, \, A=33, \, B=69
\displaystyle a	\displaystyle =	\displaystyle \frac{18}{\sin 69\degree} \times \sin 33\degree	Multiply both sides by \sin 33\degree
	\displaystyle =	\displaystyle 10.50	Evaluate

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Example 2

Consider the triangle with two interior angles C=72.53\degree and B=31.69\degree , and one side length a=5.816 metres.

a

Solve for the unknown interior angle A.

Worked Solution

Create a strategy

Subtract the given angles from 180\degree.

Apply the idea

Remember that the sum of the interior angles of a triangle is 180\degree. We are given two of the three angles of the triangle: B=31.69\degree and C=72.53\degree.

\displaystyle A+31.69\degree+72.53\degree	\displaystyle =	\displaystyle 180\degree	Add all interior angles and equate to 180\degree
\displaystyle A+104.22\degree	\displaystyle =	\displaystyle 180\degree	Evaluate the left side
\displaystyle A	\displaystyle =	\displaystyle 75.78\degree	Subtract 104.22\degree from both sides

b

Solve for b. Round your answer to three decimal places.

Worked Solution

Create a strategy

Use the sine rule: \dfrac{b}{\sin B}=\dfrac{a}{\sin A}.

Apply the idea

We are given that a=5.816 and B=31.69\degree, and we found in part (a) that A=75.78\degree.

\displaystyle \dfrac{b}{\sin 31.69\degree}	\displaystyle =	\displaystyle \dfrac{5.816}{\sin 75.78\degree}	Substitute the values
\displaystyle b	\displaystyle =	\displaystyle \dfrac{5.816 } {\sin 75.78\degree}\times\sin 31.69\degree	Multiply both sides by \sin 31.69\degree
	\displaystyle =	\displaystyle 3.152 \text{ m}	Evaluate using your calculator

c

Solve for c. Round your answer to three decimal places.

Worked Solution

Create a strategy

Use the sine rule: \dfrac{c}{\sin C}=\dfrac{a}{\sin A}.

Apply the idea

We are given that a=5.816 and C=72.53\degree, and we found in part (a) that A=75.78\degree.

\displaystyle \dfrac{c}{\sin 72.53\degree}	\displaystyle =	\displaystyle \dfrac{5.816}{\sin 75.78\degree}	Substitute the values
\displaystyle c	\displaystyle =	\displaystyle \dfrac{5.816} {\sin 75.78\degree} \times\sin 72.53\degree	Multiply both sides by \sin 72.53\degree
	\displaystyle =	\displaystyle 5.723 \text{ m}	Evaluate using your calculator

Reflect and check

We can also use B=31.69\degree and C=72.53\degree, and we found in part (b) that b=3.152.

\displaystyle \dfrac{c}{\sin 72.53\degree}	\displaystyle =	\displaystyle \dfrac{3.152}{\sin 31.69\degree}	Substitute the values
\displaystyle c	\displaystyle =	\displaystyle \dfrac{3.152 } {\sin 31.69\degree}\times\sin 72.53\degree	Multiply both sides by \sin 72.53\degree
	\displaystyle =	\displaystyle 5.723 \text{ m}	Evaluate using your calculator

We can see that we still have the same answer as the sine rule states \dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}.

Suppose we have the angles a and b, and the angle B (opposite side length b). The sine rule can be used to calculate unknown angle A (opposite side length a). Using the form of the sine rule with numerator lengths \dfrac{\sin A}{a}=\dfrac{\sin B}{b}, this equation can be transposed to make \sin A the subject of the equation, by multiplying both sides by a. This gives \sin A = \dfrac{a \sin B}{b}.

Finally, take the inverse sine of both sides of the equation to make A the subject, which gives A = \sin^{-1} \left( \dfrac{a \sin B}{b} \right).

Examples

Example 3

Find the value of the acute angle A using the Sine Rule. Write your answer in degrees to two decimal places.

Worked Solution

Create a strategy

Use the sine rule: \dfrac{\sin A}{a}=\dfrac{\sin B}{b}.

Apply the idea

\displaystyle \dfrac{\sin A}{98}	\displaystyle =	\displaystyle \dfrac{\sin 50^\circ }{81}	Substitute a=98, \, b=81, \, B=50\degree
\displaystyle \sin A	\displaystyle =	\displaystyle 98 \times\frac{\sin 50\degree }{81}	Multiply both sides by 98
\displaystyle \sin A	\displaystyle =	\displaystyle 0.926\,819	Evaluate
\displaystyle A	\displaystyle =	\displaystyle \sin ^{-1}\left(0.926\,819\right)	Take the inverse sine of both sides
	\displaystyle =	\displaystyle 67.94\degree	Evaluate

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Example 4

Find the value of the acute angle x using the sine rule. Round your answer to one decimal place.

Worked Solution

Create a strategy

Use the sine rule: \dfrac{\sin C}{c}=\dfrac{\sin A}{a}

Apply the idea

We are given two of the three side lengths of the triangle: BC=a=17 and AB=c=9, and one angle, A=56\degree. We are solving for the angle, C=x .

\displaystyle \dfrac{\sin x}{9}	\displaystyle =	\displaystyle \dfrac{\sin 56\degree}{17}	Substitute the values
\displaystyle \sin x	\displaystyle =	\displaystyle \dfrac{\sin 56\degree}{17} \times 9	Multiply both sides by 9
\displaystyle \sin x	\displaystyle =	\displaystyle 0.4389	Evaluate
\displaystyle x	\displaystyle =	\displaystyle \sin^{-1} \left(0.4389 \right)	Take the inverse sine of both sides
	\displaystyle =	\displaystyle 26.0\degree	Evaluate

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Sometimes when trying to find an unknown angle in a triangle, the sides of the triangle can be arranged in two different ways, to create either an obtuse angle or an acute angle. When this can occur, this is known as the ambiguous case of using the sine rule.

There are two cases, and what separates one from the other is summarized in this table.

This is called the ambiguous case.

This can be seen algebraically by looking at the equation used before:B = \sin^{-1} \left( \dfrac{b \sin A}{a}\right).

In the  previous lesson  , it was seen that identical (positive) trigonometric ratios can be used to find both an acute angle solution and an obtuse angle solution. The two angles that create the same sine ratio are always supplementary, meaning that their sum is 180\degree.

A calculator will only ever give the acute angle solution and not the obtuse angle. However, since the two values of B that are produced (red and green angles) always add to 180\degree, this sum can be used to find the other obtuse angle. In summary:

When finding an angle using the sine rule (with two known sides and a known angle), use the following rule to find one value of B: \dfrac{\sin B}{b}=\dfrac{\sin A}{a}

B=\sin^{-1}\left(\dfrac{b\sin{A}}{a}\right)

If the side opposite the known angle is the shorter side, this is the ambiguous case, meaning the correct angle is either acute or obtuse. Subtract the first value of B from 180\degree to find the second solution.

Examples

Example 5

Find the value of x using the sine rule, noting that x is obtuse. Round your answer to two decimal places.

Worked Solution

Create a strategy

Use the sine rule: \dfrac{\sin A}{a}=\dfrac{\sin B}{b}.

Apply the idea

\displaystyle \dfrac{\sin x}{19}	\displaystyle =	\displaystyle \dfrac{\sin 28 \degree }{14}	Substitute A=x, \,a=19, \, b=14, \, B=28
\displaystyle \sin x	\displaystyle =	\displaystyle 19 \times\frac{\sin 28\degree }{14}	Multiply both sides by 19
\displaystyle \sin x	\displaystyle =	\displaystyle 0.637\,14	Evaluate
\displaystyle x	\displaystyle =	\displaystyle \sin ^{-1}\left(0.637\,14\right)	Take the inverse sine of both sides
	\displaystyle =	\displaystyle 39.58\degree	Evaluate

To find the obtuse angle x, we subtract the acute angle from 180\degree:

\displaystyle x	\displaystyle =	\displaystyle 180 - 39.58	Subtract 39.58 from 180
	\displaystyle =	\displaystyle 140.42\degree	Evaluate

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Example 6

Consider \triangle ABC below:

a

Find x, noting that x is acute. Round your answer to the nearest degree.

Worked Solution

Create a strategy

Use the sine rule: \dfrac{\sin C}{c}=\dfrac{\sin A}{a}.

Apply the idea

\displaystyle \dfrac{\sin x}{30}	\displaystyle =	\displaystyle \dfrac{\sin 19 \degree }{15}	Substitute C=x, \, c=30,\,A=19\degree, \,a=15
\displaystyle \sin x	\displaystyle =	\displaystyle 30 \times\frac{\sin 19\degree }{15}	Multiply both sides by 30
\displaystyle \sin x	\displaystyle =	\displaystyle 0.651\,14	Evaluate
\displaystyle x	\displaystyle =	\displaystyle \sin ^{-1}\left(0.651\,14\right)	Take the inverse sine of both sides
	\displaystyle =	\displaystyle 41\degree	Evaluate

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b

Now find \angle ADB to the nearest whole degree, given that \angle ADB \gt \angle ACB.

Worked Solution

Create a strategy

To find the obtuse angle x, we subtract the acute angle from 180\degree.

Apply the idea

In part (a), we found the acute angle, \angle ACB = 41\degree. We need to find the supplementary angle of it. So,

\displaystyle \angle ADB	\displaystyle =	\displaystyle 180 - 41	Subtract 41 from 180
	\displaystyle =	\displaystyle 139\degree	Evaluate

Reflect and check

Notice that \triangle BCD is an isosceles triangle, which makes its base angles equal, \angle BCD = \angle BDC. So we can subtract \angle BDC=41\degree from 180\degree, to determine the angle, \angle ADB.

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Example 7

Consider a triangle \triangle ABC such that \angle CAB is equal to 57\degree, and side a=10 and b=11.

Select the most appropriate option to complete the sentence below: The triangle:

A

Can be either acute or obtuse.

B

Does not exist.

C

Must be a right triangle.

D

Must be an obtuse triangle.

E

Must be an acute triangle.

Worked Solution

Create a strategy

Remember that if the side opposite the known angle is the shorter side, then we are in the ambiguous case.

Apply the idea

In \triangle ABC the opposite of the known angle \angle CAB = 57\degree is the side with length a=10.

This side length is shorter than the other known side, b=11, resulting in an ambiguous case with two possible triangles and two values for the other angle.

The most appropriate option is option A.

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