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Question

(A) Given the quadratic equation $	an^2	heta - 4	an	heta + 1 = 0$.Using the quadratic formula $	an	heta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, w

Vedclass · Accepted Answer

$45^\circ, 30^\circ$

Vedclass · Answer

(A) Given the quadratic equation $	an^2	heta - 4	an	heta + 1 = 0$.Using the quadratic formula $	an	heta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, we get $	an	heta = \frac{4 \pm \sqrt{16 - 4}}{2} = \frac{4 \pm \sqrt{12}}{2} = 2 \pm \sqrt{3}$.Since $\alpha$ and $\beta$ are acute angles, $\alpha+\beta$ and $\alpha-\beta$ are also acute. We know $	an 75^\circ = 2+\sqrt{3}$ and $	an 15^\circ = 2-\sqrt{3}$.Thus, $\alpha+\beta = 75^\circ$ and $\alpha-\beta = 15^\circ$.Adding the two equations: $2\alpha = 90^\circ \implies \alpha = 45^\circ$.Subtracting the two equations: $2\beta = 60^\circ \implies \beta = 30^\circ$.

If $\alpha$ and $\beta$ are acute angles such that $\alpha+\beta$ and $\alpha-\beta$ satisfy the equation $\tan^2\theta - 4\tan\theta + 1 = 0$, then $\alpha$ and $\beta$ are respectively:

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