The goal of this page is to help students better understand when to use the Law of Sines and when to use the Law of Cosines

### Practice Problems

##### Problem 1

Can you use the Law of Sines , the Law of Cosines , or neither to solve the unknown side in triangle 1 below?

Law of Sines

** Just look at it**.
You can always immediately look at a triangle and tell whether or not you can use the Law of Sines.

You need either 2 sides and the non-included angle or, in this case, 2 angles and the non-included side.

The law of sines is all about opposite pairs.

In this case, we have a side of length 11 opposite a known angle of $$ 29^{\circ} $$ (first opposite pair) and we want to find the side opposite the known angle of $$ 118^\circ$$.

First Step $ \frac{\red x} {sin(118^{\circ})} = \frac{11}{ sin(29^{\circ})} $

##### Problem 2

Can you use the Law of Sines , the Law of Cosines , or neither to solve the unknown side in the triangle below?

Law of Cosines

Remember, the law of cosines is all about included angle (or knowing 3 sides and wanting to find an angle).

In this case, we have a side of length 20 and of 13 and the included angle of $$ 66^\circ$$.

First Step $ \red a^2 = b^2 + c^2 - 2bc \cdot cos( \angle a ) \\ \red a^2 = 20^2 + 13^2 - 2\cdot 20 \cdot 13 \cdot cos( 66 ) $

##### Problem 3

Law of Sines

** Just look at it**.
You can always immediately look at a triangle and tell whether or not you can use the Law of Sines.

You need either 2 sides and the non-included angle (like this triangle) or 2 angles and the non-included side.

Remember, the law of sines is all about opposite pairs.

In this case, we have a side of length 16 opposite a known angle of $$ 115^{\circ} $$ (first opposite pair) and we want to find the angle opposite the known side of length 32 . We can set up the proportion below and solve :

First Step $ \frac{sin(115^{\circ})}{16} = \frac{sin(\red x)}{32} $

##### Problem 4

Law of Cosines

Since you know 3 sides, and are trying to find an angle this is Law of Cosines problem.

First Step $ 8^2 = 5^2 + 6^2 -2(5)(6) \cdot cos( \red x) $

##### Problem 5

Law of Cosines

Since you know 2 sides, their included angle, and you are trying to find the side length opposite the angle, this is Law of Cosines problem.

First Step $ \red x^2 = 11^2 + 7^2 -2(11)(7) \cdot cos(50) $

##### Problem 6

Law of Sines

** Just look at it**.
You can always immediately look at a triangle and tell whether or not you can use the Law of Sines.

You need either 2 sides and the non-included angle (like this triangle) or 2 angles and the non-included side.

Since you know a side length (11) and its opposite angle (50) and want to calculate the angle measurement opposite the length of side 7, this is a Law of Sines problem

First Step $ \frac{sin ( \red x)} {7 } = \frac{sin(50)}{11} $