Exercise 45 Page 820

We want to show that we can find its area using the following formula.

Area : \frac{1}{2} b c sin A

To do so, we can use the other general formula for the area of a triangle related to its base and height. Let's also substitute the base of our triangle

b

and its height

h

into this formula.

Area : \frac{1}{2} \cdot Base \cdot Height \Rightarrow \frac{1}{2} b h

Great! Next, we will consider the left-hand side triangle with its height to write the sine ratio of

∠ A .

With the expressions for the lengths of the opposite side

h

and the hypotenuse

c,

we can write

sin A .

sin θ = \frac{Opposite side}{Hypotenuse} \Rightarrow sin A = \frac{h}{c}

Now, we will rearrange this ratio to express the height

h .

Let's multiply both sides of the equation by

c .

sin A = \frac{h}{c} \Leftrightarrow c \cdot sin A = h

Finally, we will substitute

h = c \cdot sin A

into the general area formula.

Area = \frac{1}{2} b h

Substitute

$h = c \cdot s i n A$

Area = \frac{1}{2} b \cdot (c \cdot sin A)

Multiply

Area = \frac{1}{2} b c sin A

We obtained the formula to find the area of a triangle if we have two side lengths and the measure of the included angle. Furthermore, we can also use the other side lengths and included angle combinations to find the area of a triangle because of the same relation between them.

\frac{1}{2} b c sin A = \frac{1}{2} c a sin B = \frac{1}{2} a b sin C