What do you do first if you are asked to find the area of a triangle and are given two sides and the included angle?

You are familiar with the formula R = 1 2 b h to find the area of a triangle where b is the length of a base of the triangle and h is the height, or the length of the perpendicular to the base from the opposite vertex.

Suppose Δ A B C has side lengths a , b , and c . Let h be the length of the perpendicular to the side of length b from the vertex B that meets the side A C ¯ at D .

Then, the area R of the triangle A B C is R = 1 2 b h .

Now, look at Δ A D B . It is a right triangle with hypotenuse A B ¯ that has a length of c units.

Consider the sine of ∠ A .

sin ( A ) = Opposite Side Hypotenuse = h c sin ( A ) = h c ⇒ h = c sin ( A )

Substituting the value of h in the formula for the area of a triangle, you get

R = 1 2 b ( c sin ( A ) ) = 1 2 b c sin ( A )

Similarly, you can write formulas for the area in terms of sin ( B ) or sin ( C ) .

R = 1 2 a b sin ( C ) R = 1 2 a c sin ( B )

Example 1:

Find the area of Δ P Q R .

You have the lengths of two sides and the measure of the included angle. So, you can use the formula R = 1 2 p r sin ( Q ) where p and r are the lengths of the sides opposite to the vertices P and R respectively.

Using the formula the area, R = 1 2 ( 3 ) ( 4 ) sin ( 145 ° ) .

Simplify.

R = 6 sin ( 145 ° ) ≈ 6 ( 0.5736 ) ≈ 3.44

Therefore, the area of Δ P Q R is about 3.44 sq.cm.

Example 2:

The area of the right Δ X Y Z with the right angle at the vertex Y is 39 sq. units. If Y Z = 12 and X Z = 13 , solve the triangle.

First, draw a figure with the given measures.

Use the Pythagorean Theorem to find the length of the third side of the triangle.

X Y = ( X Z ) 2 − ( Y Z ) 2 = 13 2 − 12 2 = 169 − 144 = 25 = 5

Now, you have lengths of the three sides and the area of the triangle.

Substitute in the area formula.

Area = 1 2 × ( Y Z ) × ( X Z ) × sin ( Z ) 39 = 1 2 ( 12 ) ( 13 ) sin ( Z )