7.1: Law of Sines and Law of Cosines | Precalculus

How it works

Pair each side with its opposite angle. The law of cosines handles SAS and SSS; the law of sines handles known opposite pairs. A computed inverse sine may require checking a second possible angle.

\frac{a}{\sin A}=\frac{b}{\sin B},\quad c^2=a^2+b^2-2ab\cos C

Worked examples

Example 1

Sides $\displaystyle a=5$ , $\displaystyle b=7$ enclose angle $\displaystyle C=60^\circ$ . Find c.

$\displaystyle c^2=25+49-70(1/2)=39$ , hence $\displaystyle c=\sqrt{39}$ .

Example 2

If $\displaystyle A=30^\circ$ , $\displaystyle B=45^\circ$ , and $\displaystyle a=4$ , find b.

$\displaystyle b=4\sin45^\circ/\sin30^\circ=4\sqrt{2}$ .

Problems

Problem 7.1.1

Two sides of lengths $\displaystyle 5$ and $\displaystyle 7$ enclose a $\displaystyle 60^\circ$ angle. Find the third side.

Triangle with sides 5 and 7 enclosing a 60-degree angle; opposite side c is unknown. — Match the included angle to its opposite side in the cosine rule.

Show worked solution

The law of cosines gives $\displaystyle c^2=25+49-2(5)(7)\cos60^\circ=39$ . Thus $\displaystyle c=\sqrt{39}$ .

Triangle with sides 5 and 7 enclosing a 60-degree angle; opposite side c is unknown. The solution adds labeled relationships. — The opposite side is sqrt(39).

Problem 7.1.2

For sides $\displaystyle a=3,b=4,c=5$ , find angle C using the law of cosines.

Show worked solution

$\displaystyle \cos C=(9+16-25)/(2\cdot3\cdot4)=0$ , so $\displaystyle C=90^\circ$ .

Problem 7.1.3

If $\displaystyle A=30^\circ$ , $\displaystyle B=60^\circ$ , and $\displaystyle a=6$ , find b and C.

Show worked solution

$\displaystyle C=90^\circ$ and $\displaystyle b=6\sin60^\circ/\sin30^\circ=6\sqrt{3}$ .

Lesson Problems PDF Solutions PDF Unit MCQ

How it works

Worked examples

Example 1

Example 2

Problems

Problem 7.1.1

Problem 7.1.2

Problem 7.1.3

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