Question 69

If 3sinθ+ 4cosθ = 5 (0< θ <90 ) then the value of sinθ is

Solution

Expression : $$3sin\theta + 4cos\theta=5$$

=> $$4cos\theta=5-3sin\theta$$

Squaring both sides,

=> $$(4cos\theta)^2=(5-3sin\theta)^2$$

=> $$16cos^2\theta = 25+9sin^2\theta-30sin\theta$$

=> $$16(1-sin^2\theta)=25+9sin^2\theta-30sin\theta$$

=> $$16-16sin^2\theta=25+9sin^2\theta-30sin\theta$$

=> $$25sin^2\theta-30sin\theta+9=0$$

=> $$(5sin\theta-3)^2=0$$

=> $$5sin\theta=3$$

=> $$sin\theta=\frac{3}{5}$$

=> Ans - (C)


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