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TS EAMCET 19-July-2022 Shift 1 Solved Paper

Section: Mathematics

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Question : 27 of 160

Marks: +1, -0

In a triangle

ABC

a=7, c=11, \cos A=\frac{17}{22}, \cos C=\frac{1}{14}

b \tan\frac{B}{2} \tan\frac{C-A}{2}=

18
14
2
9

Solution: 👈: Video Solution

Let

z=b \tan\frac{B}{2} \tan\frac{C-A}{2}

=b \tan\left[90-\left(\frac{C+A}{2}\right)\right] \tan\left(\frac{C-A}{2}\right)

=b \cot\left(\frac{C+A}{2}\right) \tan\left(\frac{C-A}{2}\right)

=b\left[\frac{2\cos\left(\frac{C+A}{2}\right)\sin\left(\frac{C-A}{2}\right)}{2\sin\left(\frac{C+A}{2}\right)\cos\left(\frac{C-A}{2}\right)}\right]

=b\left[\frac{\sin C-\sin A}{\sin C+\sin A}\right]

......(i)

\cos A=\frac{b^2+c^2-a^2}{2bc}

\Rightarrow\frac{17}{22}=\frac{b^2+121-49}{2\times 11\times b}\Rightarrow 17b=b^2+72

\Rightarrow b^2-17b+72=0\Rightarrow b=9\text{ or }8

\sin A=\sqrt{1-\cos^2 A}=\sqrt{1-\left(\frac{17}{22}\right)^2}=\sqrt{\frac{195}{22}}

\sin C=\sqrt{1-\cos^2 C}=\sqrt{1-\frac{1}{196}}=\frac{\sqrt{195}}{14}

From Eq. (i),

z=b\left[\frac{\sin C-\sin A}{\sin C+\sin A}\right]=b\left[\frac{\frac{\sqrt{195}}{14}-\frac{\sqrt{195}}{22}}{\frac{\sqrt{195}}{14}+\frac{\sqrt{195}}{22}}\right]

=b\left[\frac{22-14}{22+14}\right]=\frac{2b}{9}

If

b=9

, then

z=2

and if

b=8

, then

z=\frac{16}{9}

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