What is α equal to ? [NDA I 2015]

Correct Answer is : −34-3/4−43

Correct Answer is : −34-3/4−43

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NDA Math Trigonometric Ratios of Angles and Equations

For the next two(2) items that follow:

Let

\alpha

be the root of the equation

25 \cos^{2} \theta + 5 \cos \theta - 12 = c

, where

\; \frac{\pi}{2} < \alpha < \pi

[NDA I 2015]

Question : 74 of 100

Marks: +1, -0

What is

\tan \alpha

Solution:

Given:

25 \cos^{2} \theta + 5 \cos \theta - 12 = 0

, where

\frac{\pi}{2} < \alpha < \pi

To find:

\tan \alpha

25 \cos^{2} \theta + 5 \cos \theta - 12 = 0

\Rightarrow \cos \theta = \frac{-5 \pm \sqrt{5^{2} - 4(25)(-12)}}{2(25)}

\Rightarrow \cos \theta = \frac{-5 \pm 35}{50}

\Rightarrow \cos \theta = -\frac{4}{5} \text{ or } \cos \theta = \frac{3}{5}

Now,

\frac{\pi}{2} < \alpha < \pi

. So,

\cos \alpha

is negative.

\therefore \cos \alpha = -\frac{4}{5}

So,

\cos \alpha = -\frac{4}{5}

, where

\alpha

is in second quadrant.

\therefore \sin \alpha

should be positive.

\sin \alpha = \sqrt{1 - \cos^{2} \alpha}

\Rightarrow \sin \alpha = \sqrt{1 - \frac{16}{25}}

\Rightarrow \sin \alpha = \frac{3}{5}

Now,

\tan \alpha = \frac{\sin \alpha}{\cos \alpha}

\Rightarrow \tan \alpha = -\frac{3}{4}

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