Question

Solve the following triangle and determine its area.

Short answer

The value of side:$a=6$

$$\begin{gathered} m\angle B={33}^o \\ m\angle C={90}^o \end{gathered}$$

Step-by-step solution

Step 1. Given information.

Triangle is :

Step 2. Apply cosine formula to the triangle and find a.

$$\begin{gathered} \cos \left({40}^o\right)=\frac{5^2+9^2-a^2}{2\times 5\times 9} \\ \cos \left({40}^o\right)=\frac{25+81-a^2}{90} \\ 106-a^2=90\cos \left({40}^o\right) \\ 106-a^2\approx 69 \\ {a}^2\approx 106-69 \\ a\approx 37 \\ a\approx 6 \end{gathered}$$

Step 3. Apply sine formula to the triangle.

$\frac{\sin \left({40}^o\right)}{6}=\frac{\sin \left(B\right)}{5}=\frac{\sin \left(C\right)}{9}$

By taking first two fractions,

$$\begin{gathered} \frac{\sin \left({40}^o\right)}{6}=\frac{\sin B}{5} \\ \sin B=\frac{5\sin {40}^o}{6} \\ \sin B\approx \frac{3.21}{6} \\ \sin B\approx 0.53 \\ B\approx {33}^o \end{gathered}$$

By taking first and last fraction,

$$\begin{gathered} \frac{\sin \left({40}^o\right)}{6}=\frac{\sin \left(C\right)}{9} \\ \sin C=\frac{9\sin {40}^o}{6} \\ \sin C\approx 1 \\ C={90}^o \end{gathered}$$

Step 4. To find the area of the triangle.

$A=\sqrt{s(s-a)(s-b)(s-c)}$

$$\begin{gathered} s=\frac{a+b+c}{2} \\ s=\frac{6+5+9}{2} \\ s=10 \end{gathered}$$

$$\begin{gathered} A=\sqrt{10(10-6)(10-5)(10-9)} \\ A=\sqrt{10\left(4\right)\left(5\right)\left(1\right)} \\ A=\sqrt{200} \\ A=10\sqrt{2} \\ A\approx 14.14 \end{gathered}$$

So the area of the triangle is approximately 14.14 square units.