A^2+b^2=c^2
(2sqrt3)^2+b^2=16
12+b^2=16
solve for b
b=2
Now triangle area is A=1/2 bh
so
A=(1/2)(2)(2sqrt3)
A=2sqrt3
Let point is(x,y)
x=(2-0)/5=2/5
y=(-6-5)/5=-11/5
(2/5,-11/5)
yes it is always coplanar with two intersecting lines.
<h2>
Answer:</h2>
<u>x= 90°</u>.
<h2>
Step-by-step explanation:</h2>
<h3>1. Write the expression.</h3>
<h3>2. Subtract "4" from both sides of the equation.</h3>
<h3>3. Add "8sin(x)" to both sides of the equation.</h3>
<h3>4. Divide both sides by 10.</h3>
<h3>5. Apply the arcsin of sin^-1 to both sides of the equation.</h3>
<h3>6. Conclude.</h3>
<u>x= 90°</u>.
