(C) The area of a triangle is given by the formula: $\text{Area} = \frac{1}{2}ab \sin C$. Given $a = 1$,$b = 2$,and $\angle C = 60^\circ$. Substitutin

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(C) The area of a triangle is given by the formula: $\text{Area} = \frac{1}{2}ab \sin C$. Given $a = 1$,$b = 2$,and $\angle C = 60^\circ$. Substituting these values into the formula: $\text{Area} = \frac{1}{2} \times 1 \times 2 \times \sin 60^\circ$. Since $\sin 60^\circ = \frac{\sqrt{3}}{2}$,we have: $\text{Area} = 1 \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}$.

Class 11 Mathematics Trigonometrical Equations Relation between sides and angles, Solutions of triangles

Medium

The area of triangle $ABC$,in which $a = 1$,$b = 2$,and $\angle C = 60^\circ$ is

A
$\frac{1}{2}$
B
$\sqrt{3}$
C
$\frac{\sqrt{3}}{2}$
D
$\frac{3}{2}$

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Relation between sides and angles, Solutions of triangles

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The area of triangle $ABC$,in which $a = 1$,$b = 2$,and $\angle C = 60^\circ$ is

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In a $\Delta ABC$,$\angle A = 30^\circ$ and $a = 8 \text{ cm}$,then the distance of the orthocentre from vertex $A$ is equal to (with usual notations).

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