THIS IS A RIGHT & isosceles triangle: A = 90. B = 45 THEN ANGLE C =45
THEN AB = AC = 4√2
APPLY PYTHAGORAS: BC² =AB² + AC²
===> X² = (4√2)² + (4√2)² ==>16.2 +16.2 = 64
X² 64 & X =√64 ==> X = 8
Tan3A=tan(2A+A)
We know that , tan(x+y)=(tanx + tany)/(1 - (tanx)(tany))
tan(2A+A)=(tan2A+tanA)/(1 - (tan2A)(tanA))— (1)
We know that , tan2x=2tanx/(1 - tan^2x)
So, by substituting tan2A in (1),we get,
=[2tanA/(1 - tan^2A) + tanA]/1- (2tanA/(1 - tan^2A))(tanA)]
=[2tanA+tanA - tan^3A]/[1 - tan^2A - 2tan^2A]
=[3tanA - tan^3A]/[1-3tan^2A]
Therefore, tan3A= [3tanA - tan^3A]/[1-3tan^2A]
This is 16 pints.
I hope this helps
Answer:
<h2>The other solution: x = -5</h2>
Step-by-step explanation:

That's a line:
g(x)=0.5*x-1, pick up some x, and get the y.
For instace: x=0, g(0)=y=-1, x=2, g(2)=y=0.
Two pints are enough for a line.