Question

If $3sin\theta +4cos\theta =5$, then the value of $sin\theta$ is

• A. $\frac{2}{3}$
• B. $\frac{4}{5}$
• C. $\frac{3}{5}$
• D. $\frac{5}{3}$

Answer 1

The correct option is C

$\frac{3}{5}$

$Given,3sin\theta +4cos\theta =\mathrm{5.......}\left(1\right)Squaringbothsides,wehave:(3sin\theta +4cos\theta {)}^2=(5{)}^2\Rightarrow 9{\sin }^2\theta +16{\cos }^2\theta +24sin\theta cos\theta =25\Rightarrow 9(1-{\cos }^2\theta )+16(1-{\sin }^2\theta )+24sin\theta cos\theta =25\Rightarrow 9-9co{s}^2\theta +16-16{\sin }^2\theta +24sin\theta cos\theta =25\Rightarrow 16{\sin }^2\theta +9co{s}^2\theta -24sin\theta cos\theta =0\Rightarrow (4sin\theta -3cos\theta {)}^2=0\Rightarrow 4sin\theta -3cos\theta =\mathrm{0......}(2\left)Solving\right(1\left)and\right(2),sin\theta =\frac{3}{5}$