If isosceles triangle abc has a 130° angle at vertex b, which statement must be true?
a. m∠a = 15° and m∠c = 35°
b. m∠a + m∠b = 155°
c. m∠a + m∠c = 60°
d. m∠a = 20° and m∠c = 30°

The only choice that doesn't imply something absurd is 'b' .


OK.  I didn't want to have to do this, but it seems to be necessary:

-- We have an isosceles triangle.
Two sides are equal and the third one is different.
That means that two angles are equal and the third one is different.

-- All 3 angles in a triangle always add up to 180°
One angle is 130°.  If that one were one of the two equal angles,
then just those two would add up to 260° even without the 3rd angle.
So the 130° angle must be the one that's different.

-- The other two angles add up to (180 - 130) = 50°.
And they're the ones that are equal.
So  <A  and  <C  are each  25° .

-- Choice-a).  <A=15° and <C=35°.  No.  Then the triangle wouldn't have
                                                             two equal angles.

-- Choice-b).  <A + <B = 155°.  Yes.  Given <B=130, and we calculated <A=25°.

-- Choice-c).  <A+<C = 60° .  No.  They're each 25°.
                                                   If they were each 60°, then the total
                                                   for the triangle would be (130+60+60) = 250° .

-- Choice-d).  <A20=20° and <C=30°.  No.  Then the triangle wouldn't have
                                                                  two equal angles.