In -ABC - if a-b2cos2 C2-a-b2 sin2 C2 -mc2- The value of m is

(a−b)^2cos^2 C2+(a+b)^2 sin^2 C2 = a^2 +b^2 + 2ab ( sin^2 C2 nbsp;cos^2 C2) = a^2 +b^2 + 2ab ( 2 sin^2 C2 1 ) ⋯ (i) We know, sin C2 = √((s a)(s b)ab ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard IX

Mathematics

Basics of Geometry

In △ABC , if ...

Question

In $△ A B C,$ if $(a - b)^{2} {cos}^{2} \frac{C}{2} + (a + b)^{2} {sin}^{2} \frac{C}{2} = m c^{2}$ . The value of m is

Open in App

Solution

$(a - b)^{2} {cos}^{2} \frac{C}{2} + (a + b)^{2} {sin}^{2} \frac{C}{2}$
$= a^{2} + b^{2} + 2 a b ({sin}^{2} \frac{C}{2} - {cos}^{2} \frac{C}{2})$
$= a^{2} + b^{2} + 2 a b (2 {sin}^{2} \frac{C}{2} - 1) \dots (i)$
We know, $sin \frac{C}{2} = \sqrt{\frac{(s - a) (s - b)}{a b}} = \sqrt{\frac{(c + b - a) (c + a - b)}{4 a b}}$
put in equation $(i)$
$a^{2} + b^{2} + 2 a b (2 (\frac{(c + b - a) (c + a - b)}{4 a b}) - 1)$
$= a^{2} + b^{2} + 2 a b (\frac{(c + b - a) (c + a - b)}{2 a b} - 1)$
$= a^{2} + b^{2} + c^{2} - (b - a)^{2} - 2 a b$
$= c^{2}$
$∴ m = 1$

Suggest Corrections

Similar questions

Q. If in a

△ A B C

tan \frac{A}{2}, tan \frac{B}{2}, tan \frac{C}{2}

are in H.P., then the minimum value of

cot \frac{B}{2}

Q. In a

Δ A B C,

\frac{1}{1 + {tan}^{2} \frac{A}{2}} + \frac{1}{1 + {tan}^{2} \frac{B}{2}} + \frac{1}{1 + {tan}^{2} \frac{C}{2}} = k [1 + sin \frac{A}{2} sin \frac{B}{2} sin \frac{C}{2}] .

Then the value of

k

Q. In

Δ A B C

, the least value of

c o s e c \frac{A}{2} + c o s e c \frac{B}{2} + c o s e c \frac{C}{2}

Q. In a triangle

A B C,

the value of

r cot \frac{B}{2} \cdot cot \frac{C}{2}

is equal to
(where

r

is inradius )

Q. If in

△ A B C, c (a + b) cos \frac{B}{2} = b (a + c) cos \frac{C}{2},

then the triangle is

Join BYJU'S Learning Program

In △ABC, if (a−b)2cos2C2+(a+b)2sin2C2=mc2. The value of m is

(a−b)2cos2C2+(a+b)2sin2C2 =a2+b2+2ab(sin2C2−cos2C2) =a2+b2+2ab(2sin2C2−1)⋯(i) We know, sinC2=√(s−a)(s−b)ab=√(c+b−a)(c+a−b)4ab put in equation (i) a2+b2+2ab(2((c+b−a)(c+a−b)4ab)−1) =a2+b2+2ab((c+b−a)(c+a−b)2ab−1) =a2+b2+c2−(b−a)2−2ab =c2 ∴m=1

In $△ A B C,$ if $(a - b)^{2} {cos}^{2} \frac{C}{2} + (a + b)^{2} {sin}^{2} \frac{C}{2} = m c^{2}$ . The value of m is