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In right triangle BCD with angle D = 90 , we have BC = 8 and BD = 4. Find sin B.

 Apr 26, 2021
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△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 \({{\sqrt 3}\over 2}\).

 Apr 26, 2021

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