Given right angle triangle $ACB$ (C is the right angle) has an area of 224 $mm^2$, what is the length of leg b if angle A equals 31.7deg?

Here's the scenario: I have one right triangle completely defined, I need to define a similar triangle with $1/2$ the area. Specifically, I need the adjacent side length of a given angle. To solve this I imagined the theoretical half-sized triangle mirrored along the adjacent leg b making it an isosceles triangle with area equal to that of the original 448 $mm^2$. Being isosceles, the SAS formula can be:

$$area = \frac{c^2 \sin(2A)}{2}$$

This yields:

$$c = \sqrt{\frac{2 \cdot area}{\sin(2A)}}$$

So then; $b = \cos (A)*c$

Does this sound right? It feels like there should be a simpler way.

  • $\begingroup$ I don't get the part with the similar triangle. What is wrong with the given one? $\endgroup$ – mvw Jul 16 '14 at 12:54
  • $\begingroup$ I think the similar triangle is not important. I was just trying to describe the real life situation this applies to. $\endgroup$ – Joe Jul 16 '14 at 13:18

We have $\frac{a}{b}=\tan A$. Multiply both sides by $b^2$. We get $$448=ab=b^2\tan A.$$ Now calculating $b$ is straightforward.

Remark: You got the same thing, with somewhat more effort. Take your expression for $c$, multiply by $\cos A$. Bring the $\cos A$ inside the square root, and use $\sin(2A)=2\sin A\cos A$. Your expression becomes $b=\sqrt{448\cot A}$.

  • $\begingroup$ OK, now I see it. Thank you. That is much simpler than what I was trying to do. It's good to know I got the same thing, but the rearranging my formula into yours I still do not understand. $\endgroup$ – Joe Jul 16 '14 at 13:57
  • $\begingroup$ Your formula gives $c=\sqrt{448/\sin 2A}$. Multiply by $\cos A$ to get $b$. We get $b=\sqrt{448\cos^2 A/\sin 2A}$. But $\sin 2A=2\sin A\cos A$. So $b=\sqrt{448\cos A/\sin A}=\sqrt{448\cot A}$. My method gives $\sqrt{448/\tan A}$, same thing. $\endgroup$ – André Nicolas Jul 16 '14 at 14:05
  • $\begingroup$ I need to go back to school. Thanks again for your help. $\endgroup$ – Joe Jul 16 '14 at 14:52
  • $\begingroup$ You are welcome. As to going back to school, you did the problem perfectly correctly. School has a lot of negatives, including people like me telling you what to do. $\endgroup$ – André Nicolas Jul 16 '14 at 14:57

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