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The question is as follows:

The lengths of segments $PQ$ and $PR$ are 8 inches and 5 inches, respectively, and they make a 60 degree angle at $P$. Find the third side of another triangle that has a 5-inch side, an 8-inch side, and the same area as triangle $PQR$.

I found that the height of the triangle is 2.5 cm. So using the area formula for triangle, the area of triangle PQR should be $6.25 \text{ cm}^2$. But how would I find the third side of another triangle that would have both 5-inch and 8-inch sides while also maintaining the area of $6.25 \text{ cm}^2$. Any help will be greatly appreciated!

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There are a few triangle area formulas, and the most useful one here is probably $Area=\frac12 ab\sin C$, where $a$ and $b$ are lengths of two sides, and $C$ is the included angle between them. In this case, that gives us $Area=\frac12\cdot8\cdot5\sin(60^\circ)=10\sqrt3$. That's different from the area you got, of $6.25$. Are you sure you did your calculation correctly? If you got a height of $2.5cm$, was it perpendicular to the $5cm$ base?

The thing about this formula is that it's the same as $\frac12bh$, using $b$ as the base, and calculating that $h=a\sin C$. If you orient this triangle so that $5$ is the base, then the height should be $8\sin 60^\circ=4\sqrt3$.

Anyway, once that's fixed, you can solve your problem by noting that, for two triangles to have sides of $8$ and $5$, and the same areas, we must have $\sin C$ equal in both. Then you note that $\sin 120^\circ=\sin 60^\circ$...


More direct, in a way, (but more awkward in a way) is Hero's formula: $Area=\sqrt{s(s-a)(s-b)(s-c)}$, where $s=\frac{a+b+c}{2}$ is the semi-perimeter. In this case, we can write $Area=\sqrt{\frac{13+c}{2}\frac{c-3}{2}\frac{c+3}{2}\frac{13-c}{2}}$, or $16(Area)^2=(13+c)(13-c)(c+3)(c-3)$. This simplifies to a biquadratic, which can be solved in the usual way, leading to two positive solutions.

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  • $\begingroup$ I am not familiar with Hero's formula, therefore, would like for you to elaborate more on the $Area = \frac{1}{2}\sin C$ please, especially on how it can be used to solve for the third side. $\endgroup$ – geo_freak Nov 18 '17 at 16:44
  • $\begingroup$ Sure, I've made an edit. I think you made a mistake in your height calculation. $\endgroup$ – G Tony Jacobs Nov 18 '17 at 16:52
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Area of triangle when two sides and included angle is give is equal to 1/2 X product of two sides X sin theta. Now by using herons formula(its very easy and easy to understand you can see it on wikipedia) compute side of another triangle

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