I have what appears to be a $3$-sided triangle: it is two lines on a 180 degree line at the bottom. The bottom left angle is $4x-3$ the top angle is $6x + 3$ and the bottom right angle is not given, but on the angle outside of the triangle between the outside of the triangle and the line is $9x+12$. What am I suppose to do?
[Math] Finding angles of triangles
geometrytriangles
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I think I have a solution but it will be difficult to explain. I will use the following variables:
$$ \begin{split} a&=\text{base of orange triangle} \\ b&=\text{height of orange triangle} \\ c&= \text{"left" side of green triangle} \end{split} $$
Then the hypotenuse of the orange triangle is $\sqrt{a^2+b^2}$ and the hypotenuse of the green triangle is $\sqrt{a^2+b^2+c^2}$. Since the perimeters of both triangles are equal we must have:
$$ \begin{split} &&a+b=c+\sqrt{a^2+b^2+c^2} \\ &\implies& a+b-c=\sqrt{a^2+b^2+c^2} \\ &\implies& (a+b-c)^2=a^2+b^2+c^2 \\ &\implies& a^2+b^2+c^2+2ab-2ac-2bc=a^2+b^2+c^2 \\ &\implies& ab=ac+bc \\ &\implies& c=\frac{ab}{a+b} \end{split} $$
So now we have all sides in terms of $a$ and $b$. But we can relate $a$ and $b$ using the angle we are given. So:
$$ \tan(20)=\frac{b}{a} \implies b=a\tan(20) $$
Using this we get:
$$ \begin{split} a^2+b^2=a^2+a^2\tan^2(20)=a^2(1+\tan^2(20))=\frac{a^2}{\sin^2(20)} \end{split} $$
Now we can get an expression for $a^2+b^2+c^2$ purely in terms of $a$:
$$ \begin{split} a^2+b^2+c^2&= \frac{a^2}{\sin^2(20)}+\frac{a^2b^2}{\frac{a^2}{\sin^2(20)}+2ab}\\ &=\frac{a^2}{\sin^2(20)}+\frac{a^4\tan^2(20)}{\frac{a^2}{\sin^2(20)}+2a^2\tan(20)} \\ &=\frac{a^2}{\sin^2(20)}+\frac{a^4\tan^2(20)\sin^2(20)}{a^2+2a^2\tan(20)\sin^2(20)} \\ &=\frac{a^2}{\sin^2(20)}+\frac{a^2\tan^2(20)\sin^2(20)}{1+2\tan(20)\sin^2(20)} \\ &=a^2\left[ \frac{1}{\sin^2(20)}+\frac{\tan^2(20)\sin^2(20)}{1+2\tan(20)\sin^2(20)} \right] \\ &= a^2K^2 \end{split} $$
where $K^2$ is just the number in the square brackets (the only reason I made it $K^2$ instead of $K$ is because we will need to take square roots). Finally, letting $\theta$ be the smallest angle of the green triangle, we get:
$$ \begin{split} \cos(\theta)&=\frac{\sqrt{a^2+b^2}}{\sqrt{a^2+b^2+c^2}} \\ &=\frac{\frac{a}{\sin(20)}}{aK} \\ &=\frac{1}{K\sin(20)} \end{split} $$
You can work out that $K\approx 2.9262$ and then plugging this in gives us $\theta\approx 2.34^{\circ}$.
First off, the third angle is $180^o - (90^o + 5^o) = 85^o$
As to the question at hand, if you have a right angle triangle, you can use some basic trigonometric identities to come to an answer. Most notably,
$$sin \theta = (length\ of\ the\ opposite \ side) / (length\ of\ the\ hypotenuse)$$
Unfortunately, without a more complete description of the triangle, or which calculator you are using, I can't help you more than this. This, however, should be enough to get you going on finding the answer yourself.
Best Answer
I'm including a picture to help "spell things out" visually and for a visual example of "supplementary angles." I've worked the problem out on the image as well.