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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

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

difficulty advanced
 Feb 4, 2015

Best Answer 

 #6
avatar+28188 
+5

Given that C must be between 0 and pi/3, the graph below shows that there is only one solution:

 

sin(pi/3-C)+sin(C)

 

I don't know why WolframAlpha doesn't find this.

.

 Feb 5, 2015
 #1
avatar
0

try to find the angle A.......

 Feb 4, 2015
 #2
avatar+105634 
+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

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

 

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

 

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

.
 Feb 5, 2015
 #3
avatar+105634 
+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???

 Feb 5, 2015
 #4
avatar+28188 
+5

Have a look at the following for part (ii):

 

triangle 

.

 Feb 5, 2015
 #5
avatar+105634 
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.

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

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.:)

 Feb 5, 2015
 #6
avatar+28188 
+5
Best Answer

Given that C must be between 0 and pi/3, the graph below shows that there is only one solution:

 

sin(pi/3-C)+sin(C)

 

I don't know why WolframAlpha doesn't find this.

.

Alan Feb 5, 2015
 #7
avatar+104855 
0

Very impressive, Melody and Alan...!!!

 

 Feb 5, 2015
 #8
avatar+105634 
0

Thanks Alan  :)

and

Thanks Chris :)

 Feb 5, 2015
 #9
avatar+104855 
0

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

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