Part of the value of sin(u) is cut off; I suspect it should be either sin(u) = -5/13 or sin(u) = -12/13, since (5, 12, 13) is a Pythagorean triple. I'll assume -5/13.
Expand the tan expression using the angle sum identities for sin and cos :
tan(u + v) = sin(u + v) / cos(u + v)
tan(u + v) = [sin(u) cos(v) + cos(u) sin(v)] / [cos(u) cos(v) - sin(u) sin(v)]
Since both u and v are in Quadrant III, we know that each of sin(u), cos(u), sin(v), and cos(v) are negative.
Recall that for all x,
cos²(x) + sin²(x) = 1
and it follows that
cos(u) = - √(1 - sin²(u)) = -12/13
sin(v) = - √(1 - cos²(v)) = -3/5
Then putting everything together, we have
tan(u + v)
= [(-5/13) • (-4/5) + (-12/13) • (-3/5)] / [(-12/13) • (-4/5) - (-5/13) • (-3/5)]
= 56/33
(or, if sin(u) = -12/13, then tan(u + v) = -63/16)
Step-by-step explanation:
I don't get it can post a picture with it so I can understand more better then the writing
Answer:
Yes, the right angle is angle C.
Step-by-step explanation:
The given triangle has vertices at A(–2, 3), B(–3, –6), and C(2, –1).
Recall the slope formula;
The slope of the line going through AC is
The slope of the line going through BC s
The product of the slopes of AC and BC is 
This means line AC is perpendicular to BC at C.
Hence the triangle ABC is right angled at C.
