In triangle BCD, angle C = 90 degrees, CD = 3, and BD = sqrt(14) . What is tan B?
+128408 B
sqrt 14
C 3 D
BC = sqrt [ (sqrt 14)^2 - 3^2 ] = sqrt 5
tan B = CD / BC = 3 / sqrt 5 = 3sqrt (5) / 3
+2666 By the Pythagorean Theorem, \(BC= \sqrt 5 \).
Because \(\tan = { \text{opposite} \over \text{adjacent}}\), \(\tan B = {3 \over \sqrt 5 }\), which can be rationalized to \(\color{brown}\boxed{3 \sqrt 5 \over 5}\)
