In Delta ABC,if angle C = 90^circ,angle A = 3 | vedclass

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(B) Given: $\angle C = 90^\circ$,$\angle A = 30^\circ$,and $c = 20$.Since the sum of angles in a triangle is $180^\circ$,$\angle B = 180^\circ - (90^\

Vedclass · Accepted Answer

$10, 10\sqrt{3}$

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(B) Given: $\angle C = 90^\circ$,$\angle A = 30^\circ$,and $c = 20$.Since the sum of angles in a triangle is $180^\circ$,$\angle B = 180^\circ - (90^\circ + 30^\circ) = 60^\circ$.Using the sine rule: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$.$a = \frac{c \sin A}{\sin C} = \frac{20 \sin 30^\circ}{\sin 90^\circ} = \frac{20 	imes 0.5}{1} = 10$.$b = \frac{c \sin B}{\sin C} = \frac{20 \sin 60^\circ}{\sin 90^\circ} = \frac{20 	imes \frac{\sqrt{3}}{2}}{1} = 10\sqrt{3}$.Alternatively,in a $30^\circ-60^\circ-90^\circ$ triangle,the sides are in the ratio $1 : \sqrt{3} : 2$. Given the hypotenuse $c = 20$ (which corresponds to $2$ parts),the sides are $a = 10$ and $b = 10\sqrt{3}$.

In $\Delta ABC$,if $\angle C = 90^\circ$,$\angle A = 30^\circ$,and $c = 20$,then the values of $a$ and $b$ are:

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