(B) In any $\triangle ABC$,the area $\Delta$ is given by $\Delta = \frac{1}{2}bc \sin A$. From this,we have $\sin A = \frac{2\Delta}{bc} \quad (i)$. A

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(B) In any $\triangle ABC$,the area $\Delta$ is given by $\Delta = \frac{1}{2}bc \sin A$. From this,we have $\sin A = \frac{2\Delta}{bc} \quad (i)$. Also,by the sine rule,the circumradius $R$ is given by $R = \frac{a}{2 \sin A}$,which implies $\sin A = \frac{a}{2R} \quad (ii)$. Equating $(i)$ and $(ii)$,we get $\frac{2\Delta}{bc} = \frac{a}{2R}$. Solving for $\Delta$,we obtain $\Delta = \frac{abc}{4R}$.

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

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If $R$ is the circumradius of $\triangle ABC$,then the area of $\triangle ABC$ is equal to $\ldots$.

MHT CET 2019 All MHT CET

A
$\frac{abc}{R}$
B
$\frac{abc}{4R}$
C
$\frac{abc}{3R}$
D
$\frac{abc}{2R}$

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Relation between sides and angles, Solutions of triangles

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If $R$ is the circumradius of $\triangle ABC$,then the area of $\triangle ABC$ is equal to $\ldots$.

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