# Evaluate M + N

#### anemone

##### MHB POTW Director
Staff member

Given that triangle $ABC$ is congruent to triangle $CDE$, and that $\angle A=\angle B=80^{\circ}$. Suppose that $AC=1$ and $$\displaystyle DG=\frac{2\sin 10^{\circ}}{M+N\sin^2 10^{\circ}}$$.

Evaluate $M+N$.
.

#### MarkFL

Staff member
Re: Evaluate M+N

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 $$\displaystyle M+N=\frac{1}{2}\csc^3\left(10^{\circ} \right)$$

#### Opalg

##### MHB Oldtimer
Staff member
Re: Evaluate M+N

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 $M+N = -1$.

#### anemone

##### MHB POTW Director
Staff member
Re: Evaluate M+N

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 MarkFL,

Thanks for participating and since the problem doesn't mention any restriction on the variables $M$ and $N$, you got the full credit for your solution! Bravo!

But even though I am not the question setter, I should have noticed when I solved the problem that I should add the condition on both variables so that their sum is unique. I admit I overlooked this. I'm sorry...

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 $M+N = -1$.
Hi Opalg,

Thanks for participating and I must tell you when I arrived at the step where I got $2Q\sin^3 10^{\circ}+2P \sin 10^{\circ}-1=0$, I immediately thought of the triple angle formula for the function of sine and you're recent help to me in one of the simultaneous equation threads was the inspiration to me...so, I wouldn't have solved this problem if not for what you have taught me!