Question

in any triangle ABC prove that (b+c)cosB+C/2=acosB-C/2

Answer 1

HELLO DEAR,
​​

WE KNOW:- $\bold{\boxed{\large{sinX + sinY = 2sin\frac{X + Y}{2}*cos\frac{X - Y}{2} \;\;AND\;\;sin2\theta = 2sin\theta * cos\theta}}}$

And,

IN any triangle the sides a,b,c and A,B,C are the angles respectively,

the sine Formula is $\bold{\large{a/sinA = b/sinB = c/sinC = K}}$


And sum of the angles of a triangle = 180°

it means A + B + C = 180,

A = 180 - (B + C),

B + C = 180 - A


now, we have to prove,

$\bold{\large{(b + c)cos\frac{B + C}{2} = acos\frac{B - C}{2}}}$

we can also write like this,

$\bold{\large{\frac{b + c}{a} = \frac{cos\frac{B - C}{2}}{cos\frac{B + C}{2}}}}$


now, From the sine Formula,


$\bold{\large{a/sinA = b/sinB = c/sinC = K}}$


$\bold{\large{a = KsinA , b = KsinB , c = KsinC}}$


So, L.H.S, $\bold{\large{\frac{b + c}{a} = \frac{KsinB + KsinC}{KsinA}}}$


$\bold{\large{\frac{b + c}{a} = \frac{2Ksin\frac{B + C}{2}*cos\frac{B - C}{2}}{KsinA}}}$


$\bold{\large{= \frac{2sin\frac{180 - A}{2}*cos\frac{B - C}{2}}{2sin\frac{A}{2}*cos\frac{A}{2}}}}$


$\bold{\large{=\frac{cos\frac{A}{2}*cos\frac{B - C}{2}}{sin\frac{A}{2}*cos\frac{A}{2}}}}$


$\bold{\large{=\frac{cos\frac{B - C}{2}}{sin\frac{180 - (B + C)}{2}}}}$


$\bold{\large{\frac{b + c}{a}=\frac{cos\frac{B - C}{2}}{cos\frac{B + C}{2}}}}$


HENCE,


$\bold{\large{(b + c)cos\frac{B + C}{2} = acos\frac{B - C}{2}}}$


I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

Answer:

Step-by-step explanation:

arΔABC=arΔPAB+arΔPBC+arΔPCA

=12×4a+12×5a+12×6a

=2a+52a+3a=152a=3√ a24

⇒a(3√4a−152)=0

a=0  or 3√4a=152

a=303√=103–√

∴Δ=3√(103√)24=3√(300)4=753–√  sq. units.