In a - ABC a-b-cb-c-ac-a-ba-b-c-4 b2 c2 equalsA- cos 2 AB- cos 2 BC- sin 2 AD- sin 2 B

The correct option is C sin2 A We know that, 2s=a+b+c ∴(a+b+c)(b+c a)(c+a b)(a+b c)/4b^2c^2 =2s(2s 2a)(2s 2b)(2s 2c)/4b^2c^2 =4s(s a)/bc×(s b)(s c)/bc =4cos ...

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Standard VII

Mathematics

Area of a Triangle

In a Δ ABC a+...

Question

$In a Δ A B C \frac{(a + b + c) (b + c - a) (c + a - b) (a + b - c)}{4 b^{2} c^{2}} equals$

cos² A

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cos² B

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sin² A

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sin² B

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Solution

The correct option is C
sin² A

$We know that, 2 s = a + b + c ∴ \frac{(a + b + c) (b + c - a) (c + a - b) (a + b - c)}{4 b^{2} c^{2}} = \frac{2 s (2 s - 2 a) (2 s - 2 b) (2 s - 2 c)}{4 b^{2} c^{2}} = 4 \frac{s (s - a)}{b c} \times \frac{(s - b) (s - c)}{b c} = 4 c o s^{2} \frac{A}{2} \times s i n^{2} \frac{A}{2} = s i n^{2} A$

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Similar questions

Q. Prove that

\frac{(a + b + c) (b + c - a) (c + a - b) (a + b - c)}{4 b^{2} c^{2}} = {sin}^{2} A

In any $Δ A B C,$ $\frac{(a + b + c) (b + c - a) (c + a - b) (a + b - c)}{4 b^{2} c^{2}}$ is equal to

Q. For a triangle

A B C

, the expression

\frac{(a + b + c) (b + c - a) (c + a - b) (a + b - c)}{4 b^{2} c^{2}}

is equal to

Δ

A B C

the sides opposite to angles

A, B, C

are denoted by

a, b, c

respectively.

The expression

\frac{(a + b + c) (b + c - a) (c + a - b) (a + b - c)}{4 b^{2} c^{2}}

equals?

Δ

A B C

the sides opposite to angles

A, B, C

are denoted by

a, b, c

respectively.

The expression

\frac{(a + b + c) (b + c - a) (c + a - b) (a + b - c)}{4 b^{2} c^{2}}

is equal to

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In a ΔABC(a+b+c)(b+c−a)(c+a−b)(a+b−c)4b2c2equals

The correct option is C sin2 A We know that,2s=a+b+c∴(a+b+c)(b+c−a)(c+a−b)(a+b−c)4b2c2=2s(2s−2a)(2s−2b)(2s−2c)4b2c2=4s(s−a)bc×(s−b)(s−c)bc=4cos2A2×sin2A2=sin2A

$In a Δ A B C \frac{(a + b + c) (b + c - a) (c + a - b) (a + b - c)}{4 b^{2} c^{2}} equals$