If 90^{circ}

Question

(D) Given,$90^{\circ} < A < 180^{\circ}$ and $\sin A = \frac{4}{5}$.Since $90^{\circ} < A < 180^{\circ}$,we have $45^{\circ} < \frac{A}{2} < 90^{\circ

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(D) Given,$90^{\circ} Since $90^{\circ} In this interval,$	an \frac{A}{2}$ must be greater than $1$.Using the formula $\sin A = \frac{2 	an \frac{A}{2}}{1 + 	an^2 \frac{A}{2}}$,we have:$\frac{4}{5} = \frac{2 	an \frac{A}{2}}{1 + 	an^2 \frac{A}{2}}$$4(1 + 	an^2 \frac{A}{2}) = 10 	an \frac{A}{2}$$4 	an^2 \frac{A}{2} - 10 	an \frac{A}{2} + 4 = 0$$2 	an^2 \frac{A}{2} - 5 	an \frac{A}{2} + 2 = 0$$(2 	an \frac{A}{2} - 1)(	an \frac{A}{2} - 2) = 0$This gives $	an \frac{A}{2} = \frac{1}{2}$ or $	an \frac{A}{2} = 2$.Since $45^{\circ}  1$,so $	an \frac{A}{2} = 2$.

If $90^{\circ} < A < 180^{\circ}$ and $\sin A = \frac{4}{5}$,then $\tan \frac{A}{2} =$

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