Question

In a triangle $\vartriangle ABC$_{, if $a=2,b=4$and $\angle C={{60}^{0}}$ then $\angle A$ and $\angle B$} are equal to,
A. \[{{90}^{0}},{{30}^{0}}\]
B. \[{{60}^{0}},{{60}^{0}}\]
C. \[{{30}^{0}},{{90}^{0}}\]
D. \[{{60}^{0}},{{45}^{0}}\]

Answer 1

Hint: To solve this question, we will use cosine rule \[{{c}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\cos C\]. We will substitute the values of the sides $a=2,b=4$ _{and angle $\angle C={{60}^{0}}$ and find the value of $c$}. Then we will use sine Law $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$ to derive the value of the angles $\angle A$ _{and $\angle B$}.
Formula used:
Law of sine:
$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$
Cosine rule:
 \[{{c}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\cos C\]
Complete step-by-step solution:
We have been given a triangle $\vartriangle ABC$ having sides $a=2,b=4$ _{and angle $\angle C={{60}^{0}}$ and we have to find the value of angles $\angle A$ and $\angle B$}.
We will use cosine rule \[{{c}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\cos C\]and substitute the given values of sides and angles and derive the value of side length $c$.
\[\begin{align}
  & {{c}^{2}}={{2}^{2}}+{{4}^{2}}-2(2\times 4)\cos {{60}^{0}} \\
 & {{c}^{2}}=4+16-16\times \frac{1}{2} \\
 & {{c}^{2}}=12 \\
 & c=2\sqrt{3}
\end{align}\]
We will now use Law of sine $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$to find the _{angles $\angle A$ and $\angle B$}.
We will first take $\frac{a}{\sin A}=\frac{c}{\sin C}$ to find the angle $\angle A$_{.
Now we will substitute the value of the sides $a,c$ and angle $\angle C={{60}^{0}}$ in $\frac{a}{\sin A}=\frac{c}{\sin C}$.}
$\begin{align}
  & \frac{2}{\sin A}=\frac{2\sqrt{3}}{\sin {{60}^{0}}} \\
 & \frac{1}{\sin A}=\frac{\sqrt{3}}{\frac{\sqrt{3}}{2}} \\
 & \sin A=\frac{1}{2} \\
 & \sin A=\sin {{30}^{0}} \\
 & A={{30}^{0}}
\end{align}$
Now we will take $\frac{b}{\sin B}=\frac{c}{\sin C}$ to find the value of angle $\angle B$.
We will now substitute the value of the sides $b,c$ and angle $\angle C={{60}^{0}}$ _{in $\frac{b}{\sin B}=\frac{c}{\sin C}$}.
\[\begin{align}
  & \frac{4}{\sin B}=\frac{2\sqrt{3}}{\sin {{60}^{0}}} \\
 & \frac{4}{\sin B}=\frac{2\sqrt{3}}{\frac{\sqrt{3}}{2}} \\
 & \sin B=1
\end{align}\]
The value $1$ of sine is at angle $\sin {{90}^{0}}$.
$\begin{align}
  & \sin B=\sin {{90}^{0}} \\
 & B={{90}^{0}} \\
\end{align}$
The value of the angles $\angle A$ _{and $\angle B$} of triangle having sides $a=2,b=4$ _{and angle $\angle C={{60}^{0}}$}are $A={{30}^{0}}$ and \[B={{90}^{0}}\]. Hence the correct option is (C).
Note:
After deriving any one of the angles $\angle A$ _{and $\angle B$} using sine law we could have also used angle sum property to find the third angle. According to that property, the sum of all the angles is equal to ${{180}^{0}}$. As in the solution above after calculating the value of the angle $A={{30}^{0}}$ we used sine law.
Here we will find angle $\angle B$ using angle sum property.
$\begin{align}
  & A+B+C={{180}^{0}} \\
 & 30+B+60={{180}^{0}} \\
 & B={{90}^{0}}
\end{align}$