Sides of a triangle $A B C$ are in A.P.If $a <$ min ${b, c}$ ,then $cos A$ may be equal to

\frac{3 c - 4 b}{2 b}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{3 c - 4 b}{2 c}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{4 c - 3 b}{2 b}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{4 b - 3 c}{2 b}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is B $\frac{4 b - 3 c}{2 b}$
$∵$ Sides are in A.P and $a <$ min ${b, c}$
Case $I :$ If min ${b, c} = b$
$\Rightarrow a, b, c$ are in A.P.
$\Rightarrow 2 b = a + c$
$\Rightarrow a = 2 b - c$
$∴ cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c}$
$= \frac{b^{2} + c^{2} - {(2 b - c)}^{2}}{2 b c}$
$= \frac{b^{2} + c^{2} - 4 b^{2} - c^{2} + 4 b c}{2 b c}$
$= \frac{- 3 b^{2} + 4 b c}{2 b c}$
$= \frac{b (4 c - 3 b)}{2 b c} = \frac{4 c - 3 b}{2 c}$
Case $I I :$ If min ${b, c} = c$
$\Rightarrow 2 c = a + b$
$\Rightarrow a = 2 c - b$
$∴ cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c}$
$= \frac{b^{2} + c^{2} - {(2 c - b)}^{2}}{2 b c}$
$= \frac{b^{2} + c^{2} - {(4 c^{2} + b^{2} - 4 b c)}^{2}}{2 b c}$
$= \frac{- 3 c^{2} + 4 b c}{2 b c}$
$= \frac{c (4 b - 3 c)}{2 b c}$
$= \frac{(4 b - 3 c)}{2 b}$

Suggest Corrections

Similar questions

Sides of a triangle ABC are in AP, then cosA is equal to

66 a^{4} b^{2} c^{3}, 44 a^{3} b^{4} c^{2}, 24 a^{2} b^{3} c^{4}

3 a + 4 b - 2 c

- 6 a - 2 b + 3 c

a, b, c

2^{a} + 4^{b} + 8^{c} = 328

\frac{a + 2 b + 3 c}{a b c}

a, b, c

2^{a} + 4^{b} + 8^{c} = 328.

\frac{a + 2 b + 3 c}{a b c}

Join BYJU'S Learning Program