In a triangle ABC- if A - B -120- and sin A -2sin B -2sin C -2-1-32- then the value of 8 cos C is

Given : A B=120^∘ nbsp; sinA2sinB2sinC2=132 sinA2sinB2sinC2=132 ⇒ 2sinA2sinB2sinC2=116 ⇒(cosA B2 cosA+B2)sinC2=116 ⇒ nbsp;(12 sinC2)sinC2=116 (∵cos(A+B2) = ...

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Standard XII

Mathematics

Conditional Identities

In a triangle...

Question

In a triangle $A B C,$ if $A - B = 120^{\circ}$ and $sin \frac{A}{2} sin \frac{B}{2} sin \frac{C}{2} = \frac{1}{32},$ then the value of $8 cos C$ is

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Solution

Given :
$A - B = 120^{\circ}$
$sin \frac{A}{2} sin \frac{B}{2} sin \frac{C}{2} = \frac{1}{32}$

$sin \frac{A}{2} sin \frac{B}{2} sin \frac{C}{2} = \frac{1}{32} \Rightarrow 2 sin \frac{A}{2} sin \frac{B}{2} sin \frac{C}{2} = \frac{1}{16} \Rightarrow (cos \frac{A - B}{2} - cos \frac{A + B}{2}) sin \frac{C}{2} = \frac{1}{16} \Rightarrow (\frac{1}{2} - sin \frac{C}{2}) sin \frac{C}{2} = \frac{1}{16} (∵ cos (\frac{A + B}{2}) = cos (\frac{π}{2} - \frac{C}{2}) = sin \frac{C}{2}) \Rightarrow {sin}^{2} \frac{C}{2} - \frac{1}{2} sin \frac{C}{2} + \frac{1}{16} = 0 \Rightarrow {(sin \frac{C}{2} - \frac{1}{4})}^{2} = 0 \Rightarrow sin \frac{C}{2} = \frac{1}{4} \Rightarrow cos C = 1 - 2 {sin}^{2} \frac{C}{2} = 1 - \frac{1}{8} \Rightarrow cos C = \frac{7}{8} ∴ 8 cos C = 7$

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Similar questions

Q. In a

△ A B C

, If

A - B = 120^{o}

and

sin (\frac{A}{2}) sin (\frac{B}{2}) sin (\frac{C}{2}) = \frac{1}{32}

, then the value of

8 cos C

Q. In a triangle ABC,if

A - B = 120^{\circ}

and

sin \frac{A}{2} sin \frac{B}{2} sin \frac{C}{2} = \frac{1}{32}

then the value of

8 cos C

Q. In any triangle ABC prove that identities.

sin \frac{A}{2} sin \frac{B}{2} sin \frac{C}{2} \leq \frac{1}{8}

Q. If

P

is a point inside

Δ A B C

such that

∠ P B C = \frac{A}{2}, ∠ P C A = \frac{B}{2}

and

∠ P A B = \frac{C}{2}

then the value of

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

is?

Q. If

A, B

and

C

are the interior angles of a triangle

A B C

, Prove that:

\frac{cos \frac{B + C}{2} cos \frac{C + A}{2} cos \frac{A + B}{2}}{sin \frac{A}{2} sin \frac{B}{2} sin \frac{C}{2}} = 1

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In a triangle ABC, if A−B=120∘ and sinA2sinB2sinC2=132, then the value of 8cosC is

Given : A−B=120∘ sinA2sinB2sinC2=132 sinA2sinB2sinC2=132⇒2sinA2sinB2sinC2=116⇒(cosA−B2−cosA+B2)sinC2=116⇒(12−sinC2)sinC2=116(∵cos(A+B2)=cos(π2−C2)=sinC2)⇒sin2C2−12sinC2+116=0⇒(sinC2−14)2=0⇒sinC2=14⇒cosC=1−2sin2C2=1−18⇒cosC=78∴8cosC=7

In a triangle $A B C,$ if $A - B = 120^{\circ}$ and $sin \frac{A}{2} sin \frac{B}{2} sin \frac{C}{2} = \frac{1}{32},$ then the value of $8 cos C$ is