Simplify by using the law of indices:y^a(b+c)×y^b(c-a)×y^c(a-b)
1 answer:
You might be interested in
Refer to the attached diagram for help.
All radii of a circle are congruent.
Tangent segs LP and LQ are congruent. (Tangent segs drawn to a circle from the same point are congruent)
Because of this we can establish OPLQ as a kite.
OPLQ is a kite implies that OL bis. angle PLQ.
An angle bis. divides an angle into 2 congruent, parts, so angle OLP must be 30 degrees.
Since tangents form right angles with the radius, angle OPL is right.
Now we have a right triangle OPL with a 30-degree angle. The other angle must be 60 degrees because of the no-choice theorem. (180-(30+90)) = 60
We know that OP is 6 because it is a radius of circle O. How do we use that to find OL? Well, we <em>could</em> use trig, but you might recognize this as a special triangle!
The side opposite the 30 degree angle = x
The side opposite the right angle = 2x
So OL must be 12!
All radii of a circle are congruent.
Tangent segs LP and LQ are congruent. (Tangent segs drawn to a circle from the same point are congruent)
Because of this we can establish OPLQ as a kite.
OPLQ is a kite implies that OL bis. angle PLQ.
An angle bis. divides an angle into 2 congruent, parts, so angle OLP must be 30 degrees.
Since tangents form right angles with the radius, angle OPL is right.
Now we have a right triangle OPL with a 30-degree angle. The other angle must be 60 degrees because of the no-choice theorem. (180-(30+90)) = 60
We know that OP is 6 because it is a radius of circle O. How do we use that to find OL? Well, we <em>could</em> use trig, but you might recognize this as a special triangle!
The side opposite the 30 degree angle = x
The side opposite the right angle = 2x
So OL must be 12!