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# i get no diea!

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1506
9 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?

Feb 4, 2015

#6
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Given that C must be between 0 and pi/3, the graph below shows that there is only one solution: I don't know why WolframAlpha doesn't find this.

.

Feb 5, 2015

#1
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try to find the angle A.......

Feb 4, 2015
#2
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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

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

$$\\0=(2b+C)\frac{bSinA}{a}+(2c+b)\frac{cSinA}{a}-2aSinA\\\\ 0=SinA[(2b+C)\frac{b}{a}+(2c+b)\frac{c}{a}-2a]\\\\ 0=\frac{SinA}{a}[(2b+c)b+(2c+b)c-2a^2]\\\\ Neither sinA nor A can be 0 so\\\\$$

$$\\0=[2b^2+cb+2c^2+bc-2a^2]\\\\ 0=[2b^2+2c^2-2a^2+2bc]\\\\ 0=b^2+c^2-a^2+bc\\\\ -bc=b^2+c^2-a^2\\\\ -\frac{1}{2}=\frac{b^2+c^2-a^2}{2bc}\\\\ -\frac{1}{2}=cosA\qquad\qquad \mbox{Using cosine rule}\\\\ A=\pi-\frac{\pi}{3}\\\\ A=\frac{2\pi}{3}\\\\$$

Feb 5, 2015
#3
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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???

Feb 5, 2015
#4
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Have a look at the following for part (ii): .

Feb 5, 2015
#5
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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

----------------------------------------------

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

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

Thankyou.:)

Feb 5, 2015
#6
+5

Given that C must be between 0 and pi/3, the graph below shows that there is only one solution: I don't know why WolframAlpha doesn't find this.

.

Alan Feb 5, 2015
#7
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Very impressive, Melody and Alan...!!!   Feb 5, 2015
#8
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Thanks Alan  :)

and

Thanks Chris :)

Feb 5, 2015
#9
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Here's the graphical solution to the last part of Melody's answer using Desmos...

It shows that angles B and C are, indeed, equal.....

GRAPH   Feb 5, 2015