Interior/Exterior Angles and Chord of Circle

[User]

Hello. I have two homework problems that I have no idea how to go about solving.

1) An exterior angle at the base of an isosceles triangle is 130 degrees. Find the number of degrees in the vertex angle.
A friend told me the answer was 80 degrees, but I'm having trouble figuring out why. I'm sure I'm missing something really obvious, but I can't think what.

2) Find the length of a chord of a circle of radius 13, if the chord is 5 units from the center of the circle.
I have no idea how to do this, if someone could point me in the right direction that would be very helpful.

Thanks for your time.

 

[pka]

1) An exterior angle at the base of an isosceles triangle is 130 degrees. Find the number of degrees in the vertex angle. the answer is 80 degrees.

Do you see that each base angle is \(\displaystyle 50^o~?\) WHY?
\(\displaystyle 50+50=100\).

 

[pappus]

Hello. I have two homework problems that I have no idea how to go about solving.

1) An exterior angle at the base of an isosceles triangle is 130 degrees. Find the number of degrees in the vertex angle.
A friend told me the answer was 80 degrees, but I'm having trouble figuring out why. I'm sure I'm missing something really obvious, but I can't think what.

Interior and ecterior angle are together 180°. So you can determine one interior angle at the base. The other angle at the base has the same value. And all interior angles sum up to 180°.

2) Find the length of a chord of a circle of radius 13, if the chord is 5 units from the center of the circle.
I have no idea how to do this, if someone could point me in the right direction that would be very helpful.

Thanks for your time.

Draw a sketch.

Half of the chord, the part of the radius which is perpendicular to the chord has the length 5 units and the radius form a right triangle. Use Pythagorean theorem. Afterwards double the result.

 

[User]

Thanks so much, I get it now.