Problem: Triangle ABCA B CABC has a right angle at CCC. If sinA=23\sin A=\dfrac{2}{3}sinA=32, then tanB\tan BtanB is
Answer Choices:
A. 35\dfrac{3}{5}53
B. 53\dfrac{\sqrt{5}}{3}35
C. 25\dfrac{2}{\sqrt{5}}52
D. 52\dfrac{\sqrt{5}}{2}25
E. 53\dfrac{5}{3}35
Solution:
In the figure, sinA=BCAB=23\sin A=\dfrac{B C}{A B}=\dfrac{2}{3}sinA=ABBC=32. So for some x>0x>0x>0, BC=2x,AB=3xB C=2 x, A B=3 xBC=2x,AB=3x and AC=(AB)2−(BC)2=5xA C=\sqrt{(A B)^{2}-(B C)^{2}}=\sqrt{5} xAC=(AB)2−(BC)2=5x.
Thus tanB=ACBC=52\tan B=\dfrac{A C}{B C}=\dfrac{\sqrt{5}}{2}tanB=BCAC=25.