In a triangle ABC, a, b and c are the opposite sides of each angles ,and 2a*SinA=(2b+c)*SinB+(2c+b)*SinC

(I） try to the angle A

(II) if SinB+SinC=1, then what is the shape of this triangle?

Guest Feb 4, 2015

#2**+5 **

In a triangle ABC, a, b and c are the opposite sides of each angles ,and 2a*SinA=(2b+c)*SinB+(2c+b)*SinC

(I） try to the angle A

$$\\2aSinA=(2b+c)SinB+(2c+b)SinC\\

0=(2b+c)SinB+(2c+b)SinC-2aSinA\\$$

Using the sine rule I know that

$$\\SinB=\frac{bSinA}{a}\qquadand\qquad SinC=\frac{cSinA}{a}\\\\

$Substituting I get$\\\\$$

$$\\0=(2b+C)\frac{bSinA}{a}+(2c+b)\frac{cSinA}{a}-2aSinA\\\\

0=SinA[(2b+C)\frac{b}{a}+(2c+b)\frac{c}{a}-2a]\\\\

0=\frac{SinA}{a}[(2b+c)b+(2c+b)c-2a^2]\\\\

$Neither sinA nor A can be 0 so$\\\\$$

$$\\0=[2b^2+cb+2c^2+bc-2a^2]\\\\

0=[2b^2+2c^2-2a^2+2bc]\\\\

0=b^2+c^2-a^2+bc\\\\

-bc=b^2+c^2-a^2\\\\

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

-\frac{1}{2}=cosA\qquad\qquad \mbox{Using cosine rule}\\\\

A=\pi-\frac{\pi}{3}\\\\

A=\frac{2\pi}{3}\\\\$$

Melody
Feb 5, 2015

#3**+5 **

Part 2

(II) if SinB+SinC=1, then what is the shape of this triangle?

$$\\ SinB+SinC=1\\

Sin\left(\frac{\pi}{3}-C\right)+SinC=1\\$$

Now I am really confused because according to Wolfram|Alpha there is no valid solution to this. Where B and C are acute angles.

http://www.wolframalpha.com/input/?i=sinB%2Bsin%28pi%2F3-B%29%3D1

Maybe A=2pi/3 was wrong???

Melody
Feb 5, 2015

#5**0 **

Thanks Alan,

Usin Alan's pic

(II) if SinB+SinC=1, then what is the shape of this triangle?

sinB = 1/2

SinC=1/2

so statement is true

So B=C=pi/6 is one answer.

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**May I ask 2 questions please Alan**

1) How do you know that is the only answer? and

2) Why didn't Wolfram|alpha give me that answer?

Thankyou.:)

Melody
Feb 5, 2015