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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 is
A. \[\frac{{\sqrt 3 + 1}}{3}\;{\rm{c}}{{\rm{m}}^2}\]
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. \[\frac{{\sqrt 3 + 1}}{{3\sqrt 2 }}\;{\rm{c}}{{\rm{m}}^2}\]

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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
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.
Last updated date: 21st Sep 2023
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