Expert Help with Non-Right Triangle Equations

A painter needs to cover a triangular region 65 meters by 67meters by 72meters. A can of paint covers 70 square feet. How many cans will be needed?

The length of a red rectangle is 15cm more than its widthw. A blue rectangle, which is 5cm wider and 2cm shorterthan the red one, has a perimeter of 72cm. Make a sketch of therectangles expressing all dimensions in terms of w. Thenfind the dimensions of each rectagle.

1.the length of a rectangle is 3 in. more than the width. if the perimeter is 30 in., find the area of a rectangle.
2.a square has an area of $16y{d}^2$. what is the perimeterof the square?
3.The area of a triangle is $112m^2$. If the length of thebase is 16 m, what is the height?
4.What is the maximum area for a rectangle with a perimeter of 40centimeters?
5.The length of a rectangle is 2 in. more than the width. If the perimeter is 24 in., find the area of the rectangle.

Hinge Theorem. Let ABC and DEF be triangles such that AB=DE, AC=DF, and measure angel BAC smaller measure angel EDF. Then BC smaller EF.

The side of a square is 4cm longer than the side of an equilateral triangle. The perimeter of the square is 24cm more than the perimeter of the triangle. Find the lengths of a side of the square and a side of the triangle.

The following conclusion is true. Is the argument valid or invalid?
A scalene triangle has a largest angle.
A scalene triangle has a longest side.
The largest angle in a scalene triangle is opposite the longest side.
Is this argument valid or invalid?
Valid or Invalid

Solve the triangle
a = 3, b = 2 and alpha = 40 degrees

Find the perimeter of a triangle whose vertices are the points A(-2,1) B(1,3) and C(4,-3)?

Prove that if a triangle contains one right or obtuse angle, then the other two angles must be acute. (Note: You may not use the fact that the sum of the angles in a triangle is 180 degrees.)

Finding the third side of a triangle given the area
Perimeter =$\frac{(a+b+c)}{2}$
Area = $A=\sqrt{p(p-<!--\; -\; -->a)(p-<!--\; -\; -->b)(p-<!--\; -\; -->c)}$
How do I simplify the above two equations to solve for c?

the bottom or base edge b,
the top left edge a,
the top right edge c.
Let a=c+2 and b=10.

In a non right triangle $PQR$, the median from $R$ meets the sides $PQ$ at $S$, the perpendicular from $P$ meets sides $QR$ at $E$ and $RS$ and $PE$ intersect at $O$. $p=\sqrt{3},q=1$ and circumradius of $PQR$ is $1$.

Can you find the the various values of a non 45-45- 90 triangle if only given hypotenuse and right angle?

How to identify opposite and adjacent parts of right triangle?
For example, are the opposite and adjacent parts of the triangle changed if your flip the triangle another direction? I guess what I'm asking is are the opposite and adjacent parts of the triangle the same no matter what direction it's pointing?

I have three random points, O, A, B, with these I can get angles $\mathrm{\alpha <!--\; \alpha \; -->}$ and $\mathrm{\beta <!--\; \beta \; -->}$
How can I get the angle C, or better, directional vector of c which will be evenly in the middle, between the other two angles. Keep in mind that the angles might be of any value.
This old answer almost gets me were I need to be, but how can I change the final formula presented in the answer $g=\text{arctan}(2\tan <!--\; \; -->r)$ to work with non-right triangles? to work with non-right triangles?
I have found lots of results searching for this, but they don't seem to work in my case.
It looks like a good solution, but I don't know enough about trig to determine the part to change in the final formula to make it not right triangle dependant

We have the expression $13\sin <!--\; \; -->[{\tan }^{-<!--\; -\; -->1}<!--\; \; -->({\displaystyle \frac{12}{5}})]$.
Apparently the answer is 12, and I have to simplify it, and I'm assuming it means I have to show it's 12, without using a calculator.
Normally I show my own work in the questions, but in this case I have absolutely no clue how to. The only thing I know that might help is that $\tan <!--\; \; -->(x)={\displaystyle \frac{\sin <!--\; \; -->(x)}{\cos <!--\; \; -->(x)}}$

Find the Area of a segment of a circle if the central angle of the segment is ${105}^{\circ <!--\; \circ \; -->}$ degrees and the radius is $70$.

Number of distinct right triangles formed by connecting vertices of a unit cube.
Suppose we have a unit cube in ${\mathbb{R}}^3.$
We want to count the total number of distinct right triangles formed by connecting vertices of the unit cube. I can see that the total number of right triangles simply on a face of the cube will be $4$ and since there are $6$ faces we have $4\times <!--\; \times \; -->6=24$ right triangles on the faces alone. Of course there are others across the cube's diagonals. I think the total will be $48$, due a symmetry across the diagonals, but am not entirely sure of my reasoning here, so I may be wrong.

Explain this equality: $-<!--\; -\; -->{90}^{\circ <!--\; \circ \; -->}+{\tan }^{-<!--\; -\; -->1}<!--\; \; -->\frac{X_C}{R}=-<!--\; -\; -->{\tan }^{-<!--\; -\; -->1}<!--\; \; -->\frac{R}{X_C}$

If $a\ge <!--\; \ge \; -->b\ge <!--\; \ge \; -->c>0$ be real numbers such that $\mathrm{\forall <!--\; \forall \; -->}n\in <!--\; \in \; -->N$, there exists triangles of side lengths $a^n,b^n,c^n$. Then prove that those triangles must be Isosceles.