C-09-Q189mediumsingle_mcq

Solve: $\cos^2\theta - \sin^2\theta = 2 - 5\cos\theta$ , $\theta$ acute. $\theta = ?$

a $30^\circ$
b $45^\circ$
c $60^\circ$
d $75^\circ$

ব্যাখ্যা

Using $\cos^2\theta-\sin^2\theta=2\cos^2\theta-1$ gives $2\cos^2\theta+5\cos\theta-3=0$ , i.e. $(2\cos\theta-1)(\cos\theta+3)=0$ . Only $\cos\theta=\tfrac12$ is valid, so $\theta=60^\circ$ .

এই অধ্যায়ের আরও প্রশ্ন

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Trigonometric Ratio — সব প্রশ্ন দেখুন (217টি) →

Solve: $\cos^2\theta - \sin^2\theta = 2 - 5\cos\theta$ , $\theta$ acute. $\theta = ?$

30^\circ

45^\circ

60^\circ

75^\circ

ব্যাখ্যা

Using $\cos^2\theta-\sin^2\theta=2\cos^2\theta-1$ gives $2\cos^2\theta+5\cos\theta-3=0$ , i.e. $(2\cos\theta-1)(\cos\theta+3)=0$ . Only $\cos\theta=\tfrac12$ is valid, so $\theta=60^\circ$ .

Solve: cos⁡2θ−sin⁡2θ=2−5cos⁡θ\cos^2\theta - \sin^2\theta = 2 - 5\cos\thetacos2θ−sin2θ=2−5cosθ, θ\thetaθ acute. θ=?\theta = ?θ=?

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

Solve: cos⁡2θ−sin⁡2θ=2−5cos⁡θ\cos^2\theta - \sin^2\theta = 2 - 5\cos\thetacos2θ−sin2θ=2−5cosθ, θ\thetaθ acute. θ=?\theta = ?θ=?

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

Solve: $\cos^2\theta - \sin^2\theta = 2 - 5\cos\theta$ , $\theta$ acute. $\theta = ?$

Solve: $\cos^2\theta - \sin^2\theta = 2 - 5\cos\theta$ , $\theta$ acute. $\theta = ?$