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Given a triangle $ABC$, $\angle C$ is $65$ degrees, side $C$ is 10. The area of the triangle is $20.$ What is the perimeter?

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  • $\begingroup$ Can I assume size c is adjacent to angle C? Is it the hypotenuse? $\endgroup$ – Argon Apr 22 '12 at 19:48
  • $\begingroup$ size c is opposite angle C, it is not a right triangle $\endgroup$ – user8210 Apr 22 '12 at 19:49
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We can grind it out. We have $\frac{1}{2}ab\sin(65^\circ)=10$, so $ab$ is known. But from the Cosine Law, $a^2+b^2-2ab\cos(65^\circ)=100$, so $a^2+b^2$ is known. Thus $a^2+2ab+b^2$ is known, and now we know $a+b$.

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Hint: the area of a non-right triangle is given by:

$$A = \dfrac{1}{2} ab \sin C$$

where $a$ and $b$ are the side lengths and $C$ is $\angle C$.

Hint 2: Law of Cosines which is given by:

$c^2 = a^2 + b^2 -2 ab \cdot \cos C$

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Hints: You need to find $a+b+c$

You know c, and now you need to find $a+b$

Couple of formulas you could use:

Area = $\dfrac{1}{2} ab \sin C$

$c^2 = a^2 + b^2 -2 ab \cdot \cos C$

You know $c$, $\sin C$, and $\cos C$,

Can you find $a+b$?

One way is to solve for $a$, $b$ and take the sum

Another way is to use $(a+b)^2 = a^2 + b^2 + 2ab$ to find $a+b$ directly from these two above formulas

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The area tells us that the height from $C$ must be $4$. To get from that information to the lengths of the other sides, draw the circumcircle, with center $O$:

enter image description here

What is the radius of the circumcircle? Because of the Central Angle Theorem, $\angle AOB$ must be $2\times 65^\circ$, so $|AM|=|MB|=5$ must be $R\cdot \sin(65^\circ)$, or in other words $R$ is $\frac{5}{\sin(65^\circ)}$.

Now to find the exact position of $C$, we look for $\angle COM$. It must be such that $$R(\cos(65^\circ)-\cos(\angle COM)) = 4$$ and a bit of algebra then allows us to find $\angle COM$ as an arccosine. Adding and subtracting 65° gives us $\angle COA$ and $\angle COB$, and then the total perimeter becomes $$ R\left(2\sin 65^\circ+2\sin \frac{\angle COA}{2} +2\sin\frac{\angle COB}2 \right) $$

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  • $\begingroup$ +1: I had (nearly) the same idea but your picture is nicer! :-) $\endgroup$ – Raymond Manzoni Apr 22 '12 at 21:00

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