1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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 $b=\sqrt{448\cot A}$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.