Area of an Oblique Triangle
Area of an Oblique Triangle
The method that we will use to determine the area of an oblique triangle requires
knowing the measurement of two sides of the triangle and the included angle (SAS).
The formula used to calculate the area of ABC if b and c are the known sides and A
is the included angle is:
1
Area  bc sin A
2
Example 1: Determine the area of the following triangle:
Solution 1:
1
ac sin B
2
1
Area  14.113.6sin 410 
2
Area  62.90cm 2
Area 
Example 2: Determine the area of the following triangle:
The lengths of the three sides are known
but the measure of an included angle
must be determined. Choose one angle
and use the Law of Cosines to find its
measure.
Solution 2:
b2  c2  a2
cos A 
2bc
8.42  8.12  4.82
cos A 
28.4 8.1
cos A  .8313
cos 1 cos A  cos 1 .8313
1
bc sin A
2
1
Area  8.48.1sin 33.8 0 
2
Area  18.93cm 2
Area 
A  33.8 0
If another angle were chosen, there may be a slight variation in the area of the triangle.
This is the result of rounding values for the function(s) and any measurements that were
calculated.
Exercises:
1. Your backyard is a great spot to build a skating rink this winter. The ground must be
covered with sand that costs $16 per square metre. If the sides of the triangular plot of
land measure 8.24m., 7.67m. and 8.13m., what will be the cost of the sand needed to
cover this area?
2. Josh, Mary and Evan are playing frisbee in the school field. Their current positions
form a triangle with the angle at Josh equal to 44o and the angle at Mary equal to 21o.
If Mary is 15 metres from Evan, what is the area of the triangular plot formed by
Josh, Mary and Evan?
Solutions:
1.
b2  c2  a2
2bc
2
2
2

8.13  7.67   8.24
cos A 
28.137.67 
cos A  .4572
cos A 
cos 1 cos A  cos 1 .4572 
A  62.8 0
1
bc sin A
2
1
Area  8.137.67 sin 62.8 0
2
Area  27.73m 2
Area 


The cost of the sand is (28m 2 )($16 / m 2 )  $448.00
Area  28m 2
2. In a previous exercise, we solved this problem to determine the distance between
Mary and Josh. This distance was 19.57m.
1
ej sin M
2
1
Area  19.57 15sin 210 
2
Area  52.6m 2
Area 