In a - ABC- 2ac sinA-B-C2 is equal to

The correct option is B c^2+a^2 b^2 2ac sinA B+C2=2ac sin[ A+C2 B2 ] =2ac sin[ π2 B2 B2 ] [ ∵ A+C2=π2 B2 ] =2ac sin[ π2 B ] =2ac cos ...

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Definition of a Determinant

In a Δ ABC,...

Question

In a $Δ A B C, 2 a c sin \frac{A - B + C}{2}$ is equal to

a^{2} + b^{2} - c^{2}

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c^{2} + a^{2} - b^{2}

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b^{2} - a^{2} - c^{2}

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c^{2} - a^{2} - b^{2}

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Solution

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Similar questions

2 a c \sin (\frac{A - B + C}{2})

a^{2} + b^{2} - c^{2}

c^{2} + a^{2} - b^{2}

b^{2} - c^{2} - a^{2}

c^{2} - a^{2} - b^{2}

Mark the correct alternative in each of the following :

$I n a n y Δ A B C, the value of 2ac s i n (\frac{A - B + C}{2}) i s$

If a + b + c = 2s, then prove the following identities

(a) s² + (s − a)² + (s − b)² + (s − c)² = a² + b² + c²

(b) a² + b² − c² + 2ab = 4s (s − c)

(d) a² − b² − c² + 2ab = 4(s − b) (s − c)

(e) (2bc + a² − b² − c²) (2bc − a² + b² + c²) = 16s (s − a) (s − b) (s − c)

(f)

In a $△ A B C, 2 c a s i n (\frac{A - B + C}{2})$ is equal to

a + b + c = 0

\frac{1}{b^{2} + c^{2} - a^{2}} + \frac{1}{c^{2} + a^{2} - b^{2}} + \frac{1}{a^{2} + b^{2} - c^{2}}

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