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I'm a Scottish Higher maths student.

I was looking over some old textbooks, and came across a seemingly easy question, involving a circle within a triangle. I used the expected method to solve it; that is, it was in a chapter on Pythagoras' theorem, and I used this method to solve it. However, I was curious, and decided to solve it using trigonometry, the "SOH CAH TOA" method. Upon doing this, I received a different answer, and I don't understand why. This isn't urgent, I'm just curious! I've attached the question and my method below. Can anyone help?

Question:

An equilateral triangle has a circle inside it. The sides of the triangle are tangents to the circle. The radius of the circle is 2.4cm, and the distance from the centre of the circle to the vertex of the triangle is 5.1cm. Find the length of the triangle's side.

As I mentioned, I used pythag to find half of the equilateral triangles side, making a right angled triangle with half of the equilateral triangle's side, the radius of the circle, and the known length between the circles centre and the corner. This gave me a result of 4.5cm. I then realised there were multiple ways to solve the same problem using trigonometry. I used sine, using the same triangle I used for the pythag, and using the rules of the triangles, I found that the top, unknown angle, was 60 degrees, as the corner of the equilateral triangle had to be 60 degrees, and the hypotenuse of my smaller triangle cut it in half, leaving the inner angle as 30 degrees. However, when I solved this out, I got an answer of 4.42cm. This of course is not 4.5cm, my pythag result!

I've been looking at this over and over, but I just can't figure it out. I think all my working is correct, but if there are any mistakes, please just tell me. Thanks!

My pyhthag Method

My SOH CAH TOA Method

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  • $\begingroup$ Where did you get the 60 degrees from? That is not the correct angle... $\endgroup$ – user138335 Apr 6 '14 at 19:02
  • $\begingroup$ You've written $\sin x$ - that's wrong, you can't take the $\sin$ of a side. $\endgroup$ – user122283 Apr 6 '14 at 19:03
  • $\begingroup$ @user138335 $m\angle b=30$, and since $m\angle a+m\angle b+90=180$, $m\angle a=60$. $\endgroup$ – user122283 Apr 6 '14 at 19:04
  • $\begingroup$ Sorry, I mislabeled. I forgot I'd labeled the side "x" and I'm used to writing sinx for the angle. $\endgroup$ – Douglas Henderson Apr 6 '14 at 19:22
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The problem is with the numbers, not your methods. If we can inscribe a circle of radius $R$ in a equilateral triangle, the distance from the center to any vertex is $2R$. $$2.4 *2 = 4.8 \neq 5.1$$.

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  • $\begingroup$ This was perfect. I've been going nuts about this all weekend. Interesting that my maths book "tweaked" a question so that it fitted... $\endgroup$ – Douglas Henderson Apr 6 '14 at 19:36
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The question is wrong. If the radius of the circle is $2.4$ cm, then the distance from the centre to the vertex must be $4.8$ cm.

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  • $\begingroup$ Classic case of "over information". The 5.1 wasn't even required. The 2.4 is enough. $\endgroup$ – Guy Apr 6 '14 at 19:06
  • $\begingroup$ Sorry, I just wanted to make sure the question was understood, and I didn't know what was wrong, or how much to write. $\endgroup$ – Douglas Henderson Apr 6 '14 at 19:22

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