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How do you solve the triangle given \[m\angle C={{13}^{\circ }},m\angle A={{22}^{\circ }},c=9\]?

Answer
VerifiedVerified
476.7k+ views
Hint: In this problem, we have to solve the given triangle, from the given sides and angles. Here we can use the laws of sines to solve this problem, we can use the formula \[\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}\], since we have one side and two angles. We have the angle A and C and a side c, so we have to find the remaining sides a and b and the angle B using this formula.

Complete step by step solution:
We know that the given sides and angles of the triangles are,
 \[m\angle C={{13}^{\circ }},m\angle A={{22}^{\circ }},c=9\]
We have to find the remaining sides and angles to solve this triangle.
To do so, we can use the laws of sine formula, we know that laws of sine formula is,
\[\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}\]
We can now find the missing angle B.
We know that the angles in every triangle are supplementary, which means they have a sum of \[{{180}^{\circ }}\].
We can now write it as,
\[A+B+C={{180}^{\circ }}\]
We can now substitute the given angles values A, C, we get
 \[\begin{align}
  & \Rightarrow {{22}^{\circ }}+B+{{13}^{\circ }}={{180}^{\circ }} \\
 & \Rightarrow B={{180}^{\circ }}-{{35}^{\circ }}={{145}^{\circ }} \\
\end{align}\]
The angle \[B={{145}^{\circ }}\].
We can now find the side ‘a’ from the laws of sin formula,
We can now take,
\[\dfrac{\sin A}{a}=\dfrac{\sin C}{c}\]
\[\Rightarrow \dfrac{\sin {{22}^{\circ }}}{a}=\dfrac{\sin {{13}^{\circ }}}{9}\]
We can now use calculator for the sine values and solve for a, we get
\[\Rightarrow a=9\left( \dfrac{\sin {{22}^{\circ }}}{\sin {{13}^{\circ }}} \right)=9\left( \dfrac{0.374}{0.224} \right)\cong 15.0267\]
The side, \[a\cong 15.0267\]
We can now find the side ‘b’ from the laws of sin formula,
We can now take,
\[\dfrac{\sin C}{c}=\dfrac{\sin B}{b}\]
\[\Rightarrow \dfrac{\sin {{145}^{\circ }}}{b}=\dfrac{\sin {{13}^{\circ }}}{9}\]
We can now use calculator for the sine values and solve for b, we get
\[\Rightarrow b=9\left( \dfrac{\sin {{145}^{\circ }}}{\sin {{13}^{\circ }}} \right)=9\left( \dfrac{0.573}{0.224} \right)\cong 23.022\]
The side, \[b\cong 23.022\] .
Therefore, The angle \[B={{145}^{\circ }}\].
The side, \[a\cong 15.0267\].
The side, \[b\cong 23.022\] .

Note: Students make mistakes, while writing the correct law of sines formula, where we can take two angles and a side, which we are given, with which we can find the other side. We should also remember that the angles in every triangle are supplementary, which means they have a sum of \[{{180}^{\circ }}\].