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# help

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In triangle $$ABC$$ we have $$(a+b+c)(a+b-c)=3ab,$$ where $$a,b,$$ and $$c$$ are the sides opposite $$A, B,$$ and $$C$$ respectively. What is the degree measure of $$\angle C$$?

Oct 13, 2019

#1
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Am confused by question

What do you mean by sides OPPOSITE of A, B, C? is there a picture?

Oct 14, 2019
#2
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In triangle ABC we have (a+b+c)(a+b-c)=3ab, where a, b and c are

the sides opposite A, B and  C respectively.

What is the degree measure of $$\angle\ C=\angle\ \gamma$$  ?

$$\color{BrickRed}(a+b+c)(a+b-c)=3ab\\ a²+ab-ac+ab+b²-bc+ac+bc-c²-3ab=0\\ a²-ab+b²=c²$$

$$c²=a²+b²-\color{blue}ab\\ c²=a²+b²-\color{blue}2ab \cdot cos (\gamma)\ \color{red}(was\ wrong\ before:\ -2bc\cdot cos(\gamma))\\ 2ab\cdot cos(\gamma)=ab\\ cos(\gamma)= \frac{ab}{2ab}\\ cos(\gamma)=\frac{1}{2}$$

was corrected by me

$$\gamma=arc\ cos\ ( \frac{1}{2})$$

$$C=\gamma=60°$$

be inapplicable:

[If it is possible to determine the ratio ( $$\large \frac{a}{c}$$ $$|\ c\ge 2a$$ ) from equation$$(a+b+c) (a+b-c) = 3ab$$, $$\angle\ \gamma$$ can also be determined with the equation $$\gamma=arc\ cos\ ( \frac{a}{2c})$$.]

Danke CPhill! !

Oct 14, 2019
edited by asinus  Oct 14, 2019
edited by asinus  Oct 15, 2019
edited by asinus  Oct 16, 2019
#3
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( a + b + c) (a + b - c)  = 3ab

[ (a + b) + c ]  [(a + b) - c ]  = 3ab

(a + b)^2  - c^2  = 3ab

a^2 + 2ab + b^2 - c^2 = 3ab

a^2 + b^2   =  ab + c^2     (1)

Using the Law of Cosines

c^2  = [a^2 + b^2] - 2(a)(b) cos C      (2)

Sub (1)  into (2)    and we have that

c^2  = [ ab + c^2]  - (2ab) cos C

c^2  = ab + c^2 - (2ab) cos C       subtract c^2 from both sides

0   =  ab - 2ab cos C        subtract ab from both sides

-ab  = - 2ab cos C            divide both sides by -2ab

1/2  =  cos C

arccos (1/2)  = C  =  60°

Thanks to asinus for the hint of applying the Law of Cosines  !!!   Oct 15, 2019