Answer:
A) T and R
B) <SQR or <RQS
C) Q
Step-by-step explanation:
A) The sides of a angle are the end points which are T and R in this case.
B) Angle 2 has S and R and its end points. And the vertex which is Q is always in the middle, so the angle is <SQR and <RQS
Arcsin x + arcsin 2x = π/3
arcsin 2x = π/3 - arcsin x
sin[arcsin 2x] = sin[π/3 - arcsin x] (remember the left side is like sin(a-b)
2x = sinπ/3 cos(arcsin x)-cosπ/3 sin(arc sinx)
2x = √3/2 . cos(arcsin x) - (1/2)x)
but cos(arcsin x) = √(1-x²)===>2x = √3/2 .√(1-x²) - (1/2)x)
Reduce to same denominator:
(4x) = √3 .√(1-x²) - (x)===>5x = √3 .√(1-x²)
Square both sides==> 25x²=3(1-x²)
28 x² = 3 & x² = 3/28 & x =√(3/28)
Answer:
Step-by-step explanation:
we have
we know that
In this problem
substitute in the formula
