Solving SAS Triangles

"SAS" means "Side, Angle, Side"

SAS Triangle

This means we are given two sides and the included angle.

To solve an SAS triangle

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.

Because we don't know the angles facing the other two sides we can't use The Law of Sines, so we must use The Law of Cosines to find side a first:

a² = b² + c² − 2bc cosA

a2 = 52 + 72 − 2 × 5 × 7 × cos(49°)
a2 = 25 + 49 − 70 × cos(49°)
a2 = 74 − 70 × 0.6560...
a2 = 74 − 45.924... = 28.075...
a = √28.075...
a = 5.298...
a = 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

sin B / 5 = sin(49°) / 5.298...

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.

sin B = (sin(49°) × 5) / 5.298...
sin B = 0.7122...
B = sin−1(0.7122...)
B = 45.4° to one decimal place

 

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

C = 180° − 49° − 45.4°
C = 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:

r2 = p2 + q2 − 2pq cos R

r2 = 6.92 + 2.62 − 2 × 6.9 × 2.6 × cos(117°)
r2 = 47.61 + 6.76 − 35.88 × cos(117°)
r2 = 54.37 − 35.88 × (−0.4539...)
r2 = 54.37 + 16.289... = 70.659...
r = √70.659...
r = 8.405... = 8.41 to 2 decimal places

 

Now 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

sin P / 6.9 = sin(117°) / 8.405...
sin P = ( sin(117°) × 6.9 ) / 8.405...
sin P = 0.7313...
P = sin−1(0.7313...)
P = 47.0° to one decimal place

 

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

Q = 180° − 117° − 47.0°
Q = 16.0° to one decimal place