question_answer

In a triangle ABC,\[a=2,b=3\]and\[\sin A=\frac{2}{3}\]. Then\[cos\text{ }C\]is equal to:

A)  \[\frac{1}{2}\]                                  

B)  \[\frac{1}{3}\]

C)  \[\frac{2}{\sqrt{13}}\]  

D)         \[\frac{1}{\sqrt{13}}\]

E)  \[\frac{2}{3}\]

Correct Answer: E

Solution :

\[a=2,b=3,\sin A=\frac{2}{3}\Rightarrow {{\sin }^{2}}A=\frac{4}{9}\] \[\therefore \]  \[{{\cos }^{2}}A=1-{{\sin }^{2}}A=\frac{5}{9}\] \[\Rightarrow \]               \[\cos A=\frac{\sqrt{5}}{3}\] By sine rule   \[\frac{\sin A}{a}=\frac{\sin B}{b}\] \[\Rightarrow \]               \[\frac{2}{3}\times \frac{1}{2}=\frac{\sin B}{3}\] \[\Rightarrow \]               \[\sin B=1\]\[\Rightarrow \]\[B=90{}^\circ \] In\[\Delta ABC,\] \[\cos C=\frac{a}{b}=\frac{2}{3}\]