Question

If in triangle ABC, \[a=\left( 1+\sqrt{3} \right)cm\], $b=2\text{ }cm$ and $\angle C={{60}^{{}^\circ }}$, find the other two angles and third side.

Answer 1

Hint:We will use law of cosines to find the third side of the $\Delta ABC$, that is ${{c}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\cos \left( c \right)$ and then we will use sine formula to find the other two angles of the triangle $\dfrac{a}{\sin a}=\dfrac{b}{\sin b}=\dfrac{c}{\sin c}$.

Complete step-by-step answer:
It is given in the question that ABC is a triangle with sides \[a=\left( 1+\sqrt{3} \right)cm\], $b=2\text{ }cm$ and $\angle C={{60}^{{}^\circ }}$ then we have to find the other two angles and third side of the triangle.
The law of cosines say that is ABC is a triangle as follows

then ${{c}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\cos \left( c \right)$.
Now, we have \[a=\left( 1+\sqrt{3} \right)cm\], $b=2\text{ }cm$ and $\angle C={{60}^{{}^\circ }}$, therefore putting the values in the cosine formula, we get \[{{c}^{2}}={{\left( 1+\sqrt{3} \right)}^{2}}+{{2}^{2}}-2\left( 1+\sqrt{3} \right)\left( 2 \right)\cos \left( {{60}^{{}^\circ }} \right)\]
Using the general formula of ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ and putting the value $\cos {{60}^{{}^\circ }}=\dfrac{1}{2}$ we get,
 \[{{c}^{2}}=1+3+2\sqrt{3}+4-\left( 2+2\sqrt{3} \right)\left( 2 \right)\times \dfrac{1}{2}\], solving further,
\[{{c}^{2}}=8+2\sqrt{3}-2-2\sqrt{3}\left( 2 \right)\times \dfrac{1}{2}\], simplifying further, we get,
\[{{c}^{2}}=8+2\sqrt{3}-2-2\sqrt{3}\] or
\[{{c}^{2}}=6\].
Thus $c=\sqrt{6}$ is the third side of the given triangle ABC.
Now, by using the sine formula, we will find the other two angles $\angle A\text{ }and\text{ }\angle B$ of triangle ABC. According to sine formula
$\dfrac{a}{\sin a}=\dfrac{b}{\sin b}=\dfrac{c}{\sin c}$. We have \[a=\left( 1+\sqrt{3} \right)cm\], $b=2\text{ }cm$ and $\angle C={{60}^{{}^\circ }}$, also $c=\sqrt{6}$. Putting all the values in the sine formula, we get
$\dfrac{1+\sqrt{3}}{\sin a}=\dfrac{2}{\sin b}=\dfrac{\sqrt{6}}{\sin 60}$.
Now, for finding the value of angle B, we will equate the following ratio $\dfrac{2}{\sin b}=\dfrac{\sqrt{6}}{\sin 60}$. Therefore on solving this and using the value of $\sin 60=\dfrac{\sqrt{3}}{2}$, we get,
$\sin b=\dfrac{2\sin 60}{\sqrt{6}}=\dfrac{2\times \dfrac{\sqrt{3}}{2}}{\sqrt{6}}$ , solving further, we get,
$\sin b=\dfrac{\sqrt{3}}{\sqrt{6}}=\dfrac{1}{\sqrt{2}}$. Now, we know that $\sin {{45}^{{}^\circ }}=\dfrac{1}{\sqrt{2}}$, therefore, on comparing we get $\angle B={{45}^{{}^\circ }}$.
Now, we know that sum of all angles of a triangle is ${{180}^{{}^\circ }}$, that is, $\angle A+\angle B+\angle C={{180}^{{}^\circ }}$, therefore putting the values , we get
$\angle A+{{45}^{{}^\circ }}+{{60}^{{}^\circ }}={{180}^{{}^\circ }}$
$\angle A={{180}^{{}^\circ }}-{{60}^{{}^\circ }}-{{45}^{{}^\circ }}={{75}^{{}^\circ }}$.
Thus, $\angle A={{75}^{{}^\circ }}$.
Therefore, the other two angles of $\Delta ABC$ are $\angle A={{75}^{{}^\circ }}$ and $\angle B={{45}^{{}^\circ }}$ also the length of third side is equal to $c=\sqrt{6}$cm.

Note: Many students do not know about cosine formula and how to use it in mathematical problems. Also many of them do not remember this as a result they may skip such questions in examination. For remembering this we can take help of pythagoras theorem is pythagoras theorem also seem similar to cosine formula: Pythagoras theorem: ${{a}^{2}}+{{b}^{2}}={{c}^{2}}$ and cosine formula: ${{c}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\cos \left( c \right)$, only an extra term $2ab\cos c$ comes. Also, angle A can be calculated in a similar way as we calculated angle B by equating the necessary ratios.