Area of a triangle: SAS, ASA, and the Law of Sines (Part 4)

The typical way students remember the area K of a triangle is

K = \displaystyle \frac{1}{2} \times \hbox{Base} \times \hbox{Height}

However, there are other formulas for the area of a triangle which can be helpful if the height is not immediately known.

Case 1: SAS. Suppose that two sides and the angle between the sides — say, b and c and the measure of angle A — are known.


If \overline{CD} is an altitude for \triangle ABC, then \triangle ACD is a right triangle. Therefore,

\sin A = \displaystyle \frac{\hbox{opposite}}{\hbox{hypotenuse}} = \displaystyle \frac{h}{b}, or h = b \sin A.


K = \displaystyle \frac{1}{2} ch = \displaystyle \frac{1}{2} bc \sin A.

Using the same picture, one can also show that

K = \displaystyle \frac{1}{2} ac \sin B

Also, with a different but similar picture, one can show that

K = \displaystyle \frac{1}{2} ab \sin C

green lineAn important consequence of the SAS area formula is the Law of Sines. Since all three formulas must give the same area K, we have

\displaystyle \frac{1}{2} bc \sin A = \displaystyle \frac{1}{2} ac \sin B = \displaystyle \frac{1}{2} ab \sin C

Multiplying by \displaystyle \frac{2}{abc} produces the Law of Sines:

\displaystyle \frac{\sin A}{a} = \displaystyle \frac{\sin B}{b} = \displaystyle \frac{\sin C}{c}

green lineCase 2: ASA. Now suppose that we are given the measures of two angles and the length of the side in between them — say, angles A and B and side c. Naturally, we can also get the measure of angle C since the sum of the measures of the three angles must be 180^o.

From the SAS formula and the Law of Sines, we have

K = \displaystyle \frac{1}{2} bc \sin A \quad \hbox{and} \quad \displaystyle b = \frac{c \sin B}{\sin C}

Combining these, we find

K = \displaystyle \frac{1}{2} \frac{c \sin B}{\sin C} \cdot c \sin A

K = \displaystyle \frac{c^2 \sin A \sin B}{2 \sin C}

By similar reasoning, we can also find that

K = \displaystyle \frac{a^2 \sin B \sin C}{2 \sin A} ~~ and ~~ K = \displaystyle \frac{b^2 \sin A \sin C}{2 \sin B}