In right triangle BCD with angle D = 90 , we have BC = 8 and BD = 4. Find sin B.
+526 △BCD is -
By Pythagoras' theorem,
\(CD^2=BC^2-BD^2\)
\(=64-16\)
\(=48 \)
\(CD=4{\sqrt{3}}\)
⇒\(sin B={CD \over BC}\)
\(={4{\sqrt3} \over 8}\)
\(={{\sqrt 3}\over 2}\)
∴ The value of sin B is \({{\sqrt 3}\over 2}\).
