# Solving SAS Triangles

*"SAS" means "Side, Angle, Side"*

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To solve an SAS triangle

- use The Law of Cosines to calculate the unknown side,
- then use The Law of Sines to find the smaller of the other two angles,
- and then use the three angles add to 180° to find the last angle.

### Example 1

In this triangle we know:

- angle A = 49°
- b = 5
- and c = 7

To solve the triangle we need to find side **a** and angles **B** and **C**.

Use The Law of Cosines to find side **a** first:

a^{2} = b^{2} + c^{2} − 2bc cosA

^{2}= 5

^{2}+ 7

^{2}− 2 × 5 × 7 × cos(49°)

^{2}= 25 + 49 − 70 × cos(49°)

^{2}= 74 − 70 × 0.6560...

^{2}= 74 − 45.924... = 28.075...

**5.30**to 2 decimal places

Now we use the The Law of Sines to find the smaller of the other two angles.

**Why the smaller angle?** Because the inverse sine function gives answers less than 90° even for angles greater than 90°. By choosing the smaller angle (a triangle won't have two angles greater than 90°) we avoid that problem. Note: the smaller angle is the one facing the shorter side.

Choose angle B:

sin B / b = sin A / a

Did you notice that we didn't use **a = 5.30**. That number is rounded to 2 decimal places. It's much better to use the unrounded number 5.298... which should still be on our calculator from the last calculation.

^{-1}(0.7122...)

**45.4°**to one decimal place

Now we find angle C, which is easy using 'angles of a triangle add to 180°':

**85.6°**to one decimal place

Now we have completely solved the triangle i.e. we have found all its angles and sides.

### Example 2

This is also an SAS triangle.

First of all we will find **r** using The Law of Cosines:

r^{2} = p^{2} + q^{2} − 2pq cos R

^{2}= 6.9

^{2}+ 2.6

^{2}− 2 × 6.9 × 2.6 × cos(117°)

^{2}= 47.61 + 6.76 − 35.88 × cos(117°)

^{2}= 54.37 − 35.88 × (−0.4539...)

^{2}= 54.37 + 16.289... = 70.659...

**8.41**to 2 decimal places

Now for The Law of Sines.

Choose the smaller angle? We don't have to! Angle R is greater than 90°, so angles P and Q must be less than 90°.

sin P / p = sin R / r

^{-1}(0.7313...)

**47.0°**to one decimal place

Now we will find angle Q using 'angles of a triangle add to 180°':

**16.0°**to one decimal place

Mastering this skill needs lots of practice, so try these questions: