Question

in a right triangle ABC right angled at C,Cos A=1/2,Find value of Sin A.

Solve it pls​

Answer 1

Step-by-step explanation:

I hope it helps.......

Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{sin\:A=\frac{\sqrt{3}}{2}}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies cos \: A = \frac{1}{2} \\ \\ \tt: \implies \angle C = 90 \degree \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies sin \: A = ?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies cos \: A = \frac{1}{2} \\ \\ \tt \circ \: cos \: 60 \degree = \frac{1}{2} \\ \\ \tt: \implies cos \: A= cos \: 60 \degree \\ \\ \tt: \implies \angle A = 60 \degree \\ \\ \bold{For \: finding \: value : } \\ \tt: \implies sin \: A \\ \\ \tt: \implies sin \: 60 \degree \\ \\ \green{\tt: \implies \frac{ \sqrt{3} }{2} } \\ \\ \green{\tt \therefore sin \: A= \frac{ \sqrt{3} }{2} } \\ \\ \bold{Alternate \: method : } \\ \tt: \implies cos \:A = \frac{b}{h} \\ \\ \tt: \implies \frac{1}{2} = \frac{b}{h} \\ \\ \tt \circ \: b = 1 \\ \\ \tt \circ \: h = 2 \\ \\ \bold{Using \: Phythagoras \: theorem : } \\ \tt: \implies {h}^{2} = {p}^{2} + {b}^{2} \\ \\ \tt: \implies {2}^{2} = {p}^{2} + {1}^{2} \\ \\ \tt: \implies 4 - 1 = {p}^{2} \\ \\ \tt: \implies p = \sqrt{3} \\ \\ \bold{For \:finding \: value : } \\ \tt: \implies sin \: A = \frac{p}{h} \\ \\ \green{\tt: \implies sin \: A = \frac{ \sqrt{3} }{2}}$