Question

I've been struggling on this. Suppose an isosceles triangle ABC has A = 30° and b = c = 5. What is the length of a^2?

Answer 1

Any work on this so far ?

Answer 2

If you use the law of cosines you only have to plug in the values.
http://en.wikipedia.org/wiki/Law_of_cosines

Answer 3

yes you can use laws of cosine ..... Are you familiar with it ?

Answer 4

Use laws of cosine now ....

Answer 5

5^2(2 -/3) is that the answer?

Answer 6

I also want to propose a different solution. See the picture below, from which we can read that \(a = 5\sin{(15^\circ)} + 5\sin{(15^\circ)} = 10\sin{(15^\circ)}\), so \(a^2 = 100 \sin^2{(15^\circ)}\). Combine the identities \(\cos{(2x) = \cos^2{(x)} - \sin^2{(x)}}\) and \(\cos^2(x) + \sin^2(x) = 1\) to get \(\cos{(2x}) = 1 - 2\sin^2{(x)}; \sin^2(x) = \frac12(1-\cos(2x))\). Let \(x = 15^\circ\) to deduce that \(\sin^2(15^\circ) = \frac12(1-\cos(30^\circ)) = \frac12(1-\cos(30^\circ)) = \frac12(1-\frac{\sqrt3}{2})\). We can now simplify \(a^2 = 100 \sin^2{(15^\circ)} = 100 \frac12(1-\frac{\sqrt3}{2}) = 25(2 - \sqrt3)\).