- Thread starter
- Admin
- #1
- Feb 14, 2012
- 3,838
Hi MarkFL,My solution:
We find that:
\(\displaystyle \angle C=180^{\circ}-2\cdot80^{\circ}=20^{\circ}\)
The Law of Cosines gives us:
\(\displaystyle \overline{CD}^2=2\left(1-\cos\left(20^{\circ} \right) \right)\)
Triangle $CDG$ is similar to the two congruent triangles, hence, we may state:
\(\displaystyle \frac{\overline{CD}}{\overline{DG}}= \frac{1}{\overline{CD}}\implies\overline{DG}= \overline{CD}^2= 2\left(1-\cos\left(20^{\circ} \right) \right)\)
Using a double angle identity for cosine, we may write:
\(\displaystyle \overline{DG}=2\left(1-\left(1-2\sin^2\left(10^{\circ} \right) \right) \right)=4\sin^2\left(10^{\circ} \right)\)
Thus, we see that we require:
\(\displaystyle M+N\sin^2\left(10^{\circ} \right)=\frac{1}{2}\csc^3\left(10^{\circ} \right)\)
With $M=0$, we then have \(\displaystyle M+N=\frac{1}{2}\csc^3\left(10^{\circ} \right)\)
Hi Opalg,I agree with Mark as far as the line $DG = 4\sin^210^\circ$. But I wanted to express that in the form \(\displaystyle DG=\frac{2\sin 10^{\circ}}{M+N\sin^2 10^{\circ}}\), with $M$ and $N$ numerical constants (not involving $\sin10^\circ$). So I want to find $M$ and $N$ such that \(\displaystyle 4\sin^210^\circ = \frac{2\sin 10^{\circ}}{M+N\sin^2 10^{\circ}}\), or $2M\sin10^\circ + 2N\sin^310^\circ = 1$.
Remembering that $\sin3\theta = 3\sin\theta - 4\sin^3\theta$ and that $\sin30^\circ = 1/2$, I put $\theta = 10^\circ$, to get $\frac12 = 3\sin10^\circ - 4\sin^310^\circ$. Thus $6\sin10^\circ - 8\sin^310^\circ = 1$. Comparing that with the equation in the previous paragraph, you see that we can take $M=3$, $N=-4$ and hence $M+N = -1$.