In ABC with usual notations- a cos A-b cos B - c cos Ca-b-c-12- then the value of sin A-sin B-sin Csin Asin Bsin C is

A cos A+b cos B + c cos Ca+b+c=12 ⇒ acos A+b cos B+c cos C=a+b+c2 Using sine rule, we get asin A=bsin B = csin C = k (say) Now, nbsp; 2[sin Acos A+sin Bcos B ...

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Standard XIII

Mathematics

Sine Rule

In ABC with u...

Question

In $△ A B C$ with usual notations, $\frac{a cos A + b cos B + c cos C}{a + b + c} = \frac{1}{2},$ then the value of $\frac{sin A + sin B + sin C}{sin A sin B sin C}$ is

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Solution

$\frac{a cos A + b cos B + c cos C}{a + b + c} = \frac{1}{2}$
$\Rightarrow a cos A + b cos B + c cos C = \frac{a + b + c}{2}$
Using sine rule, we get
$\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C} = k (say)$

Now,
$2 [sin A cos A + sin B cos B + sin C cos C] = sin A + sin B + sin C \Rightarrow sin 2 A + sin 2 B + sin 2 C = sin A + sin B + sin C$
In a $△ A B C$ ,
$sin 2 A + sin 2 B + sin 2 C = 4 sin A sin B sin C$
So, $4 sin A sin B sin C = sin A + sin B + sin C$
$∴ \frac{sin A + sin B + sin C}{sin A sin B sin C} = 4$

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In △ABC with usual notations, acosA+bcosB+ccosCa+b+c=12, then the value of sinA+sinB+sinCsinAsinBsinC is

acosA+bcosB+ccosCa+b+c=12 ⇒acosA+bcosB+ccosC=a+b+c2 Using sine rule, we get asinA=bsinB=csinC=k (say) Now, 2[sinAcosA+sinBcosB+sinCcosC]=sinA+sinB+sinC⇒sin2A+sin2B+sin2C=sinA+sinB+sinC In a △ABC, sin2A+sin2B+sin2C=4sinAsinBsinC So, 4sinAsinBsinC=sinA+sinB+sinC ∴sinA+sinB+sinCsinAsinBsinC=4

In $△ A B C$ with usual notations, $\frac{a cos A + b cos B + c cos C}{a + b + c} = \frac{1}{2},$ then the value of $\frac{sin A + sin B + sin C}{sin A sin B sin C}$ is