In right triangle BCD with angle D = 90 , we have BC = 8 and BD = 4. Find sin B.
+37162 Need to find side DC via Pythagorean Theorem
8^2 = 4^2 + (DC)^2
DC = sqrt 48
Sin B = opposite side / hypotenuse = DC /hyp = sqrt 48 / 8 simplify as needed .........
Second method : Cos B = adjacent side / hypot = 4/8
Trig identity: sin^2 + cos^2 = 1
then sin = sqrt (1-cos^2) = sqrt ( 1 - 16/64) = sqrt 48 / 8
