In ∆ABC , with usual notations, if a, b, c are in A.P. then acos^2⁡C/2+ccos^2⁡A/2=

Correct answer: (C) 3b/2. Since a,b,c are in A.P. ⇒ 2b = a + c = a/2 2 cos^2 C/2 + ( c2 2 cos^2 A)/2 = a/2 1 + cos C + c/2 1 + cos A = 1/2 a + a cos C + c + c cos A = 1/2 a + c + b (using projection formula) = 1/2 2 b + b ( ∵ a, b, c are in A.P.) = 3b/2

Q52 / 150 · MHT CET 2018 · Mathematics

In $∆ A B C$ , with usual notations, if $a, b, c$ are in $A . P .$ then $a \cos^{2} (\frac{C}{2}) + c \cos^{2} (\frac{A}{2}) =$

A
$\frac{3 a}{2}$
B
$\frac{3 c}{2}$
Correct answer
C
$\frac{3 b}{2}$
D
$\frac{3 a b c}{2}$

Solution

Since

a, b, c

are in

A . P .

\Rightarrow

2b = a + c

= \frac{a}{2} (2 {cos}^{2} (\frac{C}{2})) + \frac{}{c} (2 {cos}^{2} (\frac{A}{2}))

= \frac{a}{2} (1 + cos C) + \frac{c}{2} (1 + cos A)

= \frac{1}{2} (a + a cos C + c + c cos A)

= \frac{1}{2} (a + c + b)

(using projection formula)

= \frac{1}{2} (2 b + b)

(

∵

a, b, c are in A.P.)

= \frac{3b}{2}

Hence, the correct answer is option: 'C'.

In ∆ABC , with usual notations, if a, b, c are in A.P. then acos2⁡C2+ccos2⁡A2=

Solution

Relevant Questions

In $∆ A B C$ , with usual notations, if $a, b, c$ are in $A . P .$ then $a \cos^{2} (\frac{C}{2}) + c \cos^{2} (\frac{A}{2}) =$