+1932 In right triangle $BCD$ with $\angle D = 90^\circ$, we have $BC = 10$ and $BD = 8$. Find $\sin B$.
+1801
In right triangle $BCD$ with $\angle D = 90^\circ$, we have $BC = 10$ and $BD = 8$. Find $\sin B$.
sin(B) = 0.6000
D is the right angle. So BD and CD are the legs and BC is the hypotenuse.
Note that this is a right triangle where one leg is 8 and the hypotenuse is 10.
We immediately recognize a Pythagoras Triplet and the other leg must be 6.
The sine is the opposite over the hypotenuse, so sin(B) = 6 / 10.
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