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I have an isosceles triangle with the two equal sides of length 'c', and the bottom of length 'a'. Both base angles of the triangle have measures of 'a', in degrees. For example, if 'a' were 50, both these angles would be 50 degrees. What I'm attempting to do is find the missing angle, 'c'. It's measurement in degrees is equal to that of both of the sides of length 'c', much the same way as 'a'. This is what I have so far: 2(a * sin(a)) + a = The perimeter. Also, a * sin(a) / 2 = c. However, this is as far as I can get, as I have no idea what the perimeter is. Does anyone know how to solve a problem like this?

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What do you mean by "Both base angles of the triangle have measures of 'a' in degrees."? –  K. Rmth Jan 3 at 17:45
    
Judging by your question, $d=a\cos(a)/2$ –  Gautam Shenoy Jan 3 at 17:50
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The way it's phrased, is the measure of the angles (in degrees) equal to the length of the base (in length units)? If that's what you're saying, isn't the missing angle just $180 - 2a$? –  MartianInvader Jan 3 at 17:52
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Note that choosing the angles of an isosceles triangle identifies it up to similarity. There will be one amongst the family of similar triangles which has a base of a given fixed length. Also given $a$ the whole triangle is known since you have the length of the base and the two base angles - you can use the sine rule to solve for $d$. –  Mark Bennet Jan 3 at 18:15
    
@MartianInvader, you are correct. In degrees, it is equal to the length of the base in units. However, I would like to find an exact measurement for it, not just 180 - 2a. I think you would need to solve for 'a' if you wanted to find the missing angle. I admit, I'm not even sure it's possible. However, I'm not very good with trigonometry, so I figured there might be a chance in solving it using so method that I don't know of. –  recursive recursion Jan 3 at 18:20
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Using sine rule, we get

$$\begin{align*} \frac{a}{\sin c^\circ} =& \frac{c}{\sin a^\circ}\\ a\sin a^{\circ} = &c\sin c^\circ\\ a\sin a^{\circ} =& (180-2a) \sin 2a^\circ\\ a\sin a^{\circ} =&2(180-2a)\sin a^\circ \cos a^\circ\\ a =& 2(180-2a)\cos a^\circ\\ \frac1{\cos a^\circ} =& \frac{360}a-4 \end{align*}$$

For the range of $a\in(0,90)$, the left hand side $\sec a$ is increasing, and the right hand side is decreasing. So the equilateral triangle is the only answer.

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thank you very much! I was going to ask for a proof of this, but you read my mind! –  recursive recursion Jan 3 at 18:52
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