equilaterat triangle has sides of 20 to nearest tenth what is altitude of triangle

Match each of the trigonometric expressions below with the equivalent non-trigonometric function from the following list. Enter the appropriate letter(A,B, C, D or E)in each blank
A . ${\displaystyle \tan \left(\arcsin \left(\frac{x}{8}\right)\right)}$
B . ${\displaystyle \cos \left(ar\sin \left(\frac{x}{8}\right)\right)}$
C. ${\displaystyle \left(\frac{1}{2}\right)\sin \left(2\arcsin \left(\frac{x}{8}\right)\right)}$
D. ${\displaystyle \sin \left(\arctan \left(\frac{x}{8}\right)\right)}$
E. ${\displaystyle \cos \left(\arctan \left(\frac{x}{8}\right)\right)}$

Find an equation of the plane. The plane through the points (2, 1, 2), (3, −8, 6), and (−2, −3, 1)

Find the point on the line $y=5x+3$ that is closet to the origin.

Proving ${\displaystyle \frac{1}{{(\tan \left(\frac{x}{2}\right)+1)}^2{\cos }^2\left(\frac{x}{2}\right)}=\frac{1}{1+\sin x}}$

Find the perimeter of the triangle.

What is the inradius of the octahedron with sidelength a?
1) Determine the height of one of two constituent square pyramids by considering a right triangle, using the fact that the height of the equilateral triangle is ${\displaystyle a\sqrt{\frac{3}{2}}}$. (Incidentally, by this you obtain the circumradius of the Octahedron.)
2) Now cut the (solid) octahedron along 4 of these latter heights, i.e. across two non-adjacent vertices and two midpoints of parallel sides of the ''square base''. (Figure 1)
Consider the resultant rhombic polygon, whose sidelength is ${\displaystyle a\sqrt{\frac{3}{2}}}$. Choose a convenient pair of adjacent corners A and B, s.t. the angle in A is obtuse. Then the insphere touches AB, say in M.
So the question becomes: Is there a still simpler proof?

How do you solve ${\displaystyle 5{\sin }^2\left(x\right)+8\sin x\cos x-3=0}$?
I have tried using compound angle, sum to product, pythag identities but nothing seems to work. I tried turning it into ${\displaystyle \sin 2x}$ but then I have ${\displaystyle {\sin }^{2}x}$ and ${\displaystyle \sin 2x}$ together.