Discussion
If the angles of a triangle are 30° and 45° and the included side is \((\sqrt{3}+1)\) cm, then the area of the triangle is:
Answer: A
The third angle is 180° - 30° - 45° = 105°. Let the included side be 'c'. We can use the formula Area = \(\frac{c^2 \sin A \sin B}{2 \sin C}\), where C is the angle opposite side c. Here C=105°.
Area = \(\frac{(\sqrt{3}+1)^2 \sin 30° \sin 45°}{2 \sin 105°}\). This is too complex. Let's try another way. Drop a perpendicular from the vertex with angle 105 to the side c. Let it be h. Let the side c be split into x and y. Then \(\tan 30 = h/x\) and \(\tan 45 = h/y\). So \(x=h\sqrt{3}\) and \(y=h\). We have \(x+y = h(\sqrt{3}+1) = \sqrt{3}+1\). This means h=1. Area = \(1/2 \times base \times height = 1/2 \times (\sqrt{3}+1) \times 1 = \frac{\sqrt{3}+1}{2}\) cm².
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