trigonometry
An isosceles triangle has an area of $27 cm^2$, and the angle between the two equal sides is $\frac{5\pi}{6}$. What is the length of the two equal sides? (Round your answer to one decimal place.)
I don't even know where to start finding the answer for this. That's what I'm looking for.
solution sketch:
Note that $h$ divides the triangle into two right triangles (h is the perpendicular bisector (altitude) from the base to the opposite vertex); the angles line $h$ forms with $a$ are right angles. This gives you two right triangles, and you only need one of them to compute the values you need.
Call the two known (marked) angles $\theta$ (they are equal).
If you know the length of the side $b$: $\sin\theta = \dfrac{h}{b} \implies h = b\sin\theta$
If $a$ is known, $\tan\theta = \dfrac{h}{a/2} \implies h = \dfrac{a}{2}\tan \theta$.
Now, using the pythorean theorem to relate your sides, we know that
$$h^2+\left(\frac{a}{2}\right)^2\,=\;b^2\tag{pythogorean theorem}$$
If the length of $b$ is unknown, using the pythagorean theorem, then knowing $a/2$ and $h$ will allow you to solve for $b$.
Knowing $\,b\,$ and $\,h\,$ will allow you to solve for $\,\dfrac{a}{2}\,$ by the pythagorean theorem Then double the value of $a/2$ to get $a$.
You have a regular $n$-gon inscribed in a circle of radius $b$, broken into $n$ isoceles triangles. In one of these triangles, the angle formed by the two radii (i.e., the two triangle sides of length $b$) is $1/n$th of the angle of the entire circle, so that this angle is $2\pi/n$. Note that
Thus, you can either use the definition of $\cos$ and the first observation to see that $$\cos\Bigl(\frac{2\pi/n}{2}\Bigr)=\cos(\pi/n)=\frac{h}{b}$$ and hence $$h=b\cdot\cos(\pi/n),$$ or you can use the law of sines and the second bullet to see that $$\frac{\sin(2\pi/n)}{a}=\frac{\sin(\frac{\pi-(2\pi/n)}{2})}{b}$$ and hence $$a=b\cdot\dfrac{\sin(2\pi/n)}{\sin(\frac{\pi-(2\pi/n)}{2})}.$$
Best Answer
Expounding on the above answer: You can use this formula for the area of a triangle for two sides and the opposite angle of opposite side, $$\text{Area}=\frac{1}{2}\text{ab}\sin \text{C}$$
Where $a$ and $b$ would be the lengths of the equal sides and $\text{C}$ is the angle between $a$ and $b$. But since $a$ and $b$ are equal in length let $a=c$ and $b=c$ so your formula becomes..$$\text{Area}=\frac{1}{2}\text{c}^2\sin \text{C}$$ Substitute the values you have and solve.