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The vertex angle of an isosceles triangle is 35 degrees. The length of the base is 10 centimeters. How many centimeters are in the perimeter?

I understand the problem as there are two sides with length 10 and one side of unknown length. I used laws of sines to find the side corresponding to the 35 degree angle. $$ \frac{\sin35^\circ}{x} = \frac{\sin72.5^\circ}{10} $$ I get 6 as the length of the side

The answer is 43.3. They get it by dropping an altitude from the vertex to the base and forming congruent right triangles. What am I doing wrong?

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  • $\begingroup$ I don't think I'd use the word "base" for one of the two sides of equal length. $\endgroup$ – Michael Hardy Oct 27 '14 at 0:27
  • $\begingroup$ Your application of the sine law is wrong. Remember it's sine of the angle divided by the side opposite the same angle. The "vertex" angle is the one opposite the "base" side. The alternative solution using congruent right triangles is fine for solving the problem, but not if you're required to use sine law. $\endgroup$ – Deepak Oct 27 '14 at 0:54
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I'm guessing what is intended is that the $35^\circ$ angle is opposite the side of length $10$, and the two equal sides of unknown length meet at that $35^\circ$ angle.

If you drop that perpendicular, you get one of the congruent halves having angles $90^\circ$ and $35^\circ/2=17.5^\circ$. The third angle would then be $72.5^\circ$. In that half, the side that meets the $72.5^\circ$ angle has length $5$. The hypotenuse would then be $5/\cos72.5^\circ$, and the height would be $5\tan72.5^\circ$.

So $\text{perimeter} = 10 + 2\dfrac{5}{\cos72.5^\circ}\approx43.255$.

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  • $\begingroup$ This does not use the law of sines as the question (the title) stipulates! Also, there's a typo with regard to the units of the perimeter (cm rather than degree measure). $\endgroup$ – Deepak Oct 27 '14 at 0:48
  • $\begingroup$ @Deepak : I've fixed the typo. At any rate, he said they get it by dropping an altitude, so "they" probably didn't use the law of sines either. $\endgroup$ – Michael Hardy Oct 27 '14 at 0:53
  • $\begingroup$ That's true enough. The OP needs to clarify. $\endgroup$ – Deepak Oct 27 '14 at 0:57
  • $\begingroup$ @Deepak : I'm assuming the OP wrote the subject line that way merely because of a surmise that the law of sines would be used. $\endgroup$ – Michael Hardy Oct 27 '14 at 1:06
  • $\begingroup$ Yes, after re-reading his question, I believe that is correct. However, I am worried he's going to come off with an impression that one can't use the sine law here, when it's just that he applied it wrongly. $\endgroup$ – Deepak Oct 27 '14 at 1:08
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The other answer does not use the law of sines, as your title states.

You're given that the side of $10$ lies opposite the vertex angle of $35^\circ$.

First find the other two angles (the "base angles"). As the triangle is isoceles, they are equal. Denote one base angle as $\theta$.

By the angle sum of the triangle, $2\theta + 35^\circ = 180^\circ$, giving $\theta = 72.5^\circ$.

Now employ the law of sines. Denote one of the sides adjoining the vertex as $x$.

So $\displaystyle \frac{x}{\sin 72.5^\circ} = \frac{10}{\sin 35^\circ}$.

Solving that gives you $x \approx 16.63$cm.

Clearly there are two such sides with length $x$, giving a perimeter of $2x + 10 \approx 2(16.63) + 10 = 43.26$cm. I'm rounding off to $2$ decimal places.

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