Class 11MathematicsTrigonometrical EquationsRelation between sides and angles, Solutions of triangles
MediumThe area of a $\Delta ABC$ is equal to
- A$\frac{1}{2}ab \sin C$
- B$\frac{1}{2}bc \sin A$
- C$\frac{1}{2}ca \sin B$
- D$bc \sin A$
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Similar Questions
If the median of $\Delta ABC$ through $A$ is perpendicular to $AB$,then
Difficult
View SolutionIf $4$ times the area of a $\triangle ABC$ is $c^2-(a-b)^2$,then $\sin C$ is equal to:
In any triangle $ABC$,$\frac{\tan \frac{A}{2} - \tan \frac{B}{2}}{\tan \frac{A}{2} + \tan \frac{B}{2}} = $
Medium
View SolutionIn triangle $ABC$,if $a=4, b=3, c=2$,then $2(a-b \cos C)(a-c \sec B) = $
In a $\Delta ABC$,if $b + c = 3a$,then the value of $\cot\, \frac{B}{2} \cdot \cot\, \frac{C}{2}$ is equal to:
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