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# trig

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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