In a triangle ABC, value of angle C?

Hello, fardeen_gen!

Quote:

In a triangle ABC: . $a = 2,\;b = 3,\; \tan A = \sqrt{\frac{3}{5}}$
then the value of ${\color{red}c}$ is:

$A)\;\sqrt{10} \qquad B)\;\frac{\sqrt{10}}{2} \qquad C)\;\frac{3\sqrt{10}}{2}\qquad D)\; \frac{\sqrt{10}}{3}$

Since $\tan A \:=\:\frac{\sqrt{3}}{\sqrt{5}} \:=\:\frac{opp}{adj} \quad\Rightarrow\quad \text{then: }\:hyp \:=\:\sqrt{8} \:=\:2\sqrt{2}$

. . Hence: . $\begin{array}{ccccccc}\sin A \:=\: \dfrac{\sqrt{3}}{2\sqrt{2}} &\Rightarrow & \boxed{\sin A \:=\:\dfrac{\sqrt{6}}{4}} \\ \\[-3mm]\cos A \:=\: \dfrac{\sqrt{5}}{2\sqrt{2}} &\Rightarrow& \boxed{\cos A \:=\: \dfrac{\sqrt{10}}{4}} \end{array}$

Law of Sines: . $\frac{\sin B}{b} \:=\:\frac{\sin A}{a} \quad\Rightarrow\quad \sin B \:=\:\frac{b\!\cdot\!\sin A}{a}$

. . Hence: . $\sin B \:=\:\dfrac{3\cdot\frac{\sqrt{6}}{4}}{2} \quad\Rightarrow\quad \begin{array}{ccc}\boxed{\sin B \:=\: \dfrac{3\sqrt{6}}{8}} \\ \\[-3mm] \boxed{\cos B \:=\: \dfrac{\sqrt{10}}{8}} \end{array}$

Since: . $C \:=\:180^o - (A + B)$ ,
. . we have: . $\sin C \;=\;\sin\left[180^o-(A + B)\right] \:=\:\sin(A + B)$

$\sin C \;=\;\sin A\cos B + \sin B\cos A \;=\;\left(\frac{\sqrt{6}}{4}\right)\left(\frac{\s qrt{10}}{8}\right) + \left(\frac{3\sqrt{6}}{8}\right)\left(\frac{\sqrt{ 10}}{4}\right)$

. . . . $= \;\frac{\sqrt{60}}{32} + \frac{3\sqrt{60}}{32} \;=\;\frac{4\sqrt{60}}{32} \;=\;\frac{\sqrt{60}}{8} \quad \Rightarrow\quad\boxed{\sin C \:=\:\frac{\sqrt{15}}{4}}$

Law of Sines: . $\frac{c}{\sin C} \:=\:\frac{a}{\sin A} \quad\Rightarrow\quad c \:=\:\frac{a\!\cdot\!\sin C}{\sin A}$

. . $c \;=\;\frac{2\cdot\dfrac{\sqrt{15}}{4}}{\dfrac{\sqr t{6}}{4}} \;=\;{\color{blue}\sqrt{10}} \quad\hdots \quad \text{answer A}$