Question

If in a $\Delta ABC,AB=4,BC=2,CA=3$ and $A,B,C$ are respective angles then the value of : $\left|\begin{array}{ccc}4& 3\sin A& 4\sin A\\ 3\sin A& 1& \cos A\\ 4\sin A& \cos A& 1\end{array}\right|$

Answer 1

Area of ${\Delta }^{\prime \prime }=\sqrt{s(s-a)(s-b)(s-c)}$

$=\sqrt{\frac{9}{2}(\frac{9}{2}-2)(\frac{9}{2}-3)(\frac{9}{2}-4)}$

$=\sqrt{\frac{9}{2}\times \frac{5}{2}\times \frac{3}{2}\times \frac{1}{2}}$

$=\frac{3}{4}\sqrt{15}$

$R=\frac{abc}{4\Delta }=\frac{24}{3\sqrt{15}}=\frac{8}{\sqrt{15}}=\frac{8\sqrt{15}}{15}$

$sinA=\frac{a}{2R}=\frac{2\times 15}{16\sqrt{15}}=\frac{30}{16\sqrt{15}}=\frac{2\sqrt{15}}{16}=\frac{\sqrt{15}}{8}$

$sinB=\frac{b}{2R}=\frac{3\times 15}{8\sqrt{15}}=\frac{3\sqrt{15}}{8}$

$sinC=\frac{C}{2R}=\frac{4\sqrt{15}}{8}=\frac{\sqrt{15}}{2}$

$cosA=\frac{b^2+c^2-a^2}{2bc}=\frac{21}{24}=\frac{7}{8}$

$cosB=\frac{4+16-9}{16}=\frac{11}{16}$

$cosC=\frac{4+9-16}{12}=\frac{-3}{12}=\frac{-1}{4}$

$=4(1-\frac{49}{64})-\frac{3\sqrt{5}}{8}(\frac{3\sqrt{15}}{8}-\frac{7\sqrt{15}}{16})+\frac{\sqrt{15}}{2}(\frac{21\sqrt{15}}{64}-\frac{\sqrt{15}}{2})$

$=\frac{15}{16}+\frac{3\times 15}{128}-\frac{11\times 15}{128}$

$=\frac{15}{16}-\frac{8\times 15}{128}=0$