(B) Given $A = 30^{\circ}$,$a = 7$,and $b = 8$. Calculate $b \sin A = 8 \sin 30^{\circ} = 8 \times 0.5 = 4$. Since $b \sin A < a < b$ (i.e.,$4 < 7 < 8

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(B) Given $A = 30^{\circ}$,$a = 7$,and $b = 8$. Calculate $b \sin A = 8 \sin 30^{\circ} = 8 \times 0.5 = 4$. Since $b \sin A Therefore,there are two possible values for angle $B$,which means there are two solutions.

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

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If $A = 30^{\circ}$,$a = 7$,and $b = 8$ in $\Delta ABC$,then $B$ has

A
One solution
B
Two solutions
C
No solution
D
None of these

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Trigonometrical Equations

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Relation between sides and angles, Solutions of triangles

If in a triangle $ABC$,$a = 5$,$b = 4$,and $A = \frac{\pi}{2} + B$,then $C$ is:

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In any $\triangle ABC$,the simplified form of $\frac{\cos 2A}{a^2} - \frac{\cos 2B}{b^2}$ is

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If $S$ is the circumcentre of a triangle $ABC$, $a=5, b=6, c=9$ and $SB=\frac{27}{4 \sqrt{2}}$, then $\sin 2C=$

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The lengths of the sides of a triangle are $\alpha - \beta$,$\alpha + \beta$,and $\sqrt{3\alpha^2 + \beta^2}$,where $\alpha > \beta > 0$. Its largest angle is:

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If in a $\triangle ABC$,$r_3 = r_1 + r_2 + r$,then $\angle A + \angle B$ is equal to (in $^{\circ}$)

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If $A = 30^{\circ}$,$a = 7$,and $b = 8$ in $\Delta ABC$,then $B$ has

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If in a triangle $ABC$,$a = 5$,$b = 4$,and $A = \frac{\pi}{2} + B$,then $C$ is:

In any $\triangle ABC$,the simplified form of $\frac{\cos 2A}{a^2} - \frac{\cos 2B}{b^2}$ is

If $S$ is the circumcentre of a triangle $ABC$, $a=5, b=6, c=9$ and $SB=\frac{27}{4 \sqrt{2}}$, then $\sin 2C=$

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