Wow ! There's so much extra mush here that the likelihood of being
distracted and led astray is almost unavoidable.
The circle ' O ' is roughly 98.17% (π/3.2) useless to us. The only reason
we need it at all is in order to recall that the tangent to a circle is
perpendicular to the radius drawn to the tangent point. And now
we can discard Circle - ' O ' .
Just keep the point at its center, and call it point - O .
-- The segments LP, LQ, and LO, along with the radii OP and OQ, form
two right triangles, reposing romantically hypotenuse-to-hypotenuse.
The length of segment LO ... their common hypotenuse ... is the answer
to the question.
-- Angle PLQ is 60 degrees. The common hypotenuse is its bisector.
So the acute angle of each triangle at point ' L ' is 30 degrees, and the
acute angle of each triangle at point ' O ' is 60 degrees.
-- The leg of each triangle opposite the 30-degree angle is a radius
of the discarded circle, and measures 6 .
-- In every 30-60 right triangle, the length of the side opposite the hypotenuse
is one-half the length of the hypotenuse.
-- So the length of the hypotenuse (segment LO) is <em>12 </em>.
Answer:
Step-by-step explanation:
hope this helps
Yo mean sqrt 27 or √27
remember
√(xy)=(√x)(√y)
√27=(√9)(√3)=(3)(√3)=3√3
Answer:
A) 15.8
Step-by-step explanation:
10/8 = .80
15+ .80 = 15.8
