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Properties of Triangles

Section: Mathematics

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Question : 50 of 50

Marks: +1, -0

In

\triangle ABC, \cos A+\cos B+\cos C=

[AP EAPCET 18 May 2024 Shift 1]

$\;1+\frac{r}{2R}$
$\;1-\frac{r}{R}$
$\;1+\frac{R}{r}$
$\;1+\frac{r}{R}$

Solution:

We have,

A B C

is a triangle

\;\cos A+\cos B+\cos C

\;=2\cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)+1-2\sin^2 \frac{C}{2}

\;=1+2\sin\frac{C}{2}\cos\left(\frac{A-B}{2}\right)-2\sin^2\frac{C}{2}

\;=1+2\sin\frac{C}{2}\left[\cos\left(\frac{A-B}{2}\right)-\sin\frac{C}{2}\right]

\;=1+2\sin\frac{C}{2}\left[\cos\left(\frac{A-B}{2}\right)-\cos\left(\frac{A+B}{2}\right)\right]

\;=1+2\sin\frac{C}{2}\left[2\sin\frac{A}{2}\sin\frac{B}{2}\right]

\;=1+4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}

\;=1+4\left[\sqrt{\frac{(s-b)(s-c)}{bc}} \sqrt{\frac{(s-a)(s-c)}{ac}}\right]

\;=\frac{1+4(s-a)(s-b)(s-c)}{abc}

\;=\frac{1+4 s(s-a)(s-b)(s-c)}{sabc}\left[\therefore \text{ multiply and divide by } s\right]

\;=1+4\frac{\Delta^2}{sabc},\;\text{ where } \Delta \text{ is area of triangle }

\;=1+\left(\frac{\Delta}{s}\right)\left(\frac{4\Delta}{abc}\right)\;\;\left[\because r=\frac{\Delta}{s}\text{ and }R=\frac{abc}{4\Delta}\right]

\;=1+\frac{r}{R}

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