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# If in a triangle ${\rm{ABC}}$ side $a = (\sqrt 3 + 1){\rm{cm}}$ and $\angle B = {30^\circ }$,$\angle C = {45^\circ }$ then the area of the triangle isA. $\frac{{\sqrt 3 + 1}}{3}\;{\rm{c}}{{\rm{m}}^2}$B. $\frac{{\sqrt 3 + 1}}{2}\;{\rm{c}}{{\rm{m}}^2}$C. $\frac{{\sqrt 3 + 1}}{{2\sqrt 2 }}\;{\rm{c}}{{\rm{m}}^2}$D. $\frac{{\sqrt 3 + 1}}{{3\sqrt 2 }}\;{\rm{c}}{{\rm{m}}^2}$

Last updated date: 22nd Mar 2023
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Hint:
In this case, we have been given that in a triangle ABC, the side is $a = (\sqrt 3 + 1){\rm{cm}}$ and two angles $\angle B = {30^\circ }$,$\angle C = {45^\circ }$and here we see that one of the angles is $\angle C = {45^\circ }$and the other is $\angle B = {30^\circ }$then it may be isosceles triangle. We know that all angle’s sum is ${180^\circ }$ in a triangle. So that we have to apply sine rule and also to use area of triangle formula in order to get the required result.
Formula used:
Sine rule:
$\frac{{\rm{a}}}{{\sin {\rm{A}}}} = \frac{{\rm{b}}}{{\sin {\rm{B}}}} = \frac{{\rm{c}}}{{\sin {\rm{C}}}}$
Triangle’s area:
$\frac{1}{2}{\rm{ab}}\sin {\rm{C}}$
Complete step-by-step solution:
We have been provided in the question that in a triangle $ABC$ we have
$a = (\sqrt 3 + 1);\angle B = {30^\circ },\angle C = {45^\circ }$
As we know by sum of angle property that triangle’s sum of angle is ${180^0}$ we can write as
$\Rightarrow \angle {\rm{A}} = {180^\circ } - {75^\circ } = {105^\circ }$
Now, we have to use sine rule
$\frac{{\rm{a}}}{{\sin {\rm{A}}}} = \frac{{\rm{b}}}{{\sin {\rm{B}}}} = \frac{{\rm{c}}}{{\sin {\rm{C}}}}$
Now, we have to substitute the values provided in the data to the above equation, we get
$\Rightarrow \frac{{\sqrt 3 + 1}}{{\sin {{105}^\circ }}} = \frac{{\rm{b}}}{{\sin {{30}^\circ }}} = \frac{{\rm{c}}}{{\sin {{45}^\circ }}}$
From the above expression, we have to determine the value of ${\rm{b}}$ we obtain
$\Rightarrow {\rm{b}} = \frac{{(\sqrt 3 + 1)\sin {{30}^\circ }}}{{\sin {{105}^\circ }}}$
Now, let’s solve numerator and denominator by using trigonometry identity, we get
$\frac{{(\sqrt 3 + 1)}}{{2\sin \left( {{{45}^\circ } + {{60}^\circ }} \right)}}$
On further simplification using trigonometry identity, we obtain
$= \sqrt 2$
We have been already known that the area of triangle is $\frac{1}{2}{\rm{ab}}\sin {\rm{C}}$
By using the above formula we have ti substitute the values obtain previously we get
$= \frac{1}{2}(\sqrt 3 + 1)(\sqrt 2 )\frac{1}{{\sqrt 2 }}$
Now, we have to simplify further we obtain
$= \frac{{\sqrt 3 + 1}}{2}\;{\rm{c}}{{\rm{m}}^2}$
Therefore, if in a triangle ${\rm{ABC}}$ side $a = (\sqrt 3 + 1){\rm{cm}}$ and $\angle B = {30^\circ },\angle C = {45^\circ }$ then the area of the triangle is $\frac{{\sqrt 3 + 1}}{2}\;{\rm{c}}{{\rm{m}}^2}$
Hence, the option B is correct
Note:
It is important to remember that the sum of all the angles must equal 180. Also, keep in mind that the length of a triangle's sides is a positive number. To relate the length and angles of a triangle, the sine and cosine formulas must be used.