Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively.

If $\frac{sinA}{4}=\frac{sinB}{5}=\frac{sinC}{6}$,

then value of $cosA+cosB+cosC$ is equal to?

• A. $\frac{4}{15}$
• B. $\frac{1}{13}$
• C. $\frac{2}{15}$
• D. $\frac{23}{16}$

Answer 1

The correct option is D $\frac{23}{16}$

Using sine rule, we get
$\frac{sinA}{a}=\frac{sinB}{b}=\frac{sinC}{c}$
And it is given that
$\frac{sinA}{4}=\frac{sinB}{5}=\frac{sinC}{6}$
Comparing both the expressions, we get
$a=4$, $b=5$, $c=6$.
Hence
$cosA=\frac{b^2+c^2-a^2}{2bc}$
$=\frac{25+36-16}{60}$
$=\frac{3}{4}$.
Similarly $cosB=\frac{9}{16}$
$cosC=\frac{1}{8}$
Hence
$\sum cos\theta$
$=\frac{3}{4}+\frac{1}{8}+\frac{9}{16}$
$\frac{12+2+9}{16}$
$=\frac{23}{16}$.