How do you solve the triangle given A=80, a=9, b=9.1?

Hint: To find the value of c, we will use sine law of triangles. We know that the law of sines states that the ratio of side length to the sine of the opposite angle. We can demonstrate it as $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}$ . First, we have to find the angle B from $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}$ . Then we will use the rule that sum of all angles of a triangle is ${{180}^{{}^\circ }}$ and find the angle C. Then we will use $\dfrac{b}{\sin B}=\dfrac{c}{\sin C}$ and find the value of c.

Complete step-by-step solution:
We are given that $\angle A=80,a=9,b=9.1$ . Let us illustrate this.

We will be using the sine law of triangles to find the value of side c. We know that the law of sines states that the ratio of side length to the sine of the opposite angle. We can denote this in the form of equation as
$\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}$
Let us consider $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}$ . We have to find angle B.
$\Rightarrow \sin B=\dfrac{b}{a}\sin A$
Let us substitute the values.
$\begin{align}
  & \Rightarrow \sin B=\dfrac{9.1}{9}\sin 80 \\
 & \Rightarrow \sin B=\dfrac{9.1}{9}\times 0.984 \\
 & \Rightarrow \sin B=1.011\times 0.984 \\
 & \Rightarrow \sin B=0.994 \\
\end{align}$
We have to take the inverse of sine to get the value of B.
$\begin{align}
  & \Rightarrow B={{\sin }^{-1}}\left( 0.994 \right) \\
 & \Rightarrow B={{83.72}^{{}^\circ }} \\
\end{align}$
We know that the sum of all angles of a triangle is ${{180}^{{}^\circ }}$ .
$\Rightarrow \angle A+\angle B+\angle C={{180}^{{}^\circ }}$
Let us find angle C from the above equation.
$\begin{align}
  & \Rightarrow {{80}^{{}^\circ }}+{{83.72}^{{}^\circ }}+\angle C={{180}^{{}^\circ }} \\
 & \Rightarrow {{163.72}^{{}^\circ }}+\angle C={{180}^{{}^\circ }} \\
 & \Rightarrow \angle C={{180}^{{}^\circ }}-{{163.72}^{{}^\circ }} \\
 & \Rightarrow \angle C={{16.28}^{{}^\circ }} \\
\end{align}$
Now, let us consider the second part of the sine formula.
$\Rightarrow \dfrac{b}{\sin B}=\dfrac{c}{\sin C}$
Let us rearrange the above equation.
$\Rightarrow c=\dfrac{b\sin C}{\sin B}$
Let us substitute the values in the above formula.
\[\begin{align}
  & \Rightarrow c=\dfrac{9.1\times \sin {{16.28}^{{}^\circ }}}{\sin {{83.72}^{{}^\circ }}} \\
 & \Rightarrow c=\dfrac{9.1\times 0.28}{0.993} \\
 & \Rightarrow c=\dfrac{2.548}{0.993} \\
 & \Rightarrow c=2.565 \\
\end{align}\]
Hence, the value of c is 2.565.

Note: Students must know the law of sines to solve these kinds of problems. We can also write the law of sines as $\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}$ . Students must know the common triangle laws and rules like the sum of angles of a triangle is always ${{180}^{{}^\circ }}$ .