##40^circ,40^circand100^circ##

In an Isoceles triangle, the two base angles are equal. Let the base angles be ##x##. Then, the obtuse angle would be ##2 1/2*x##

We need to find the value of ##x##

Since the sum of the angles in a triangle is ##180^circ##,

##rarrx+x+2 1/2*x=180^circ##

##rarr2x+2 1/2*x=180^circ##

##rarr2x+5/2*x=180^circ##

##rarr2x+(5x)/2=180^circ##

##rarr(9x)/2=180^circ##

##rarr9x=360^circ##

##color(green)(rArrx=40^circ##

Now we know that the base angles are ##color(green)(40^circ##

So, the other angle is ##color(orange)(2 1/2*40=100)##

##40^circ,40^circand100^circ##

In an Isoceles triangle, the two base angles are equal. Let the base angles be ##x##. Then, the obtuse angle would be ##2 1/2*x##

We need to find the value of ##x##

Since the sum of the angles in a triangle is ##180^circ##,

##rarrx+x+2 1/2*x=180^circ##

##rarr2x+2 1/2*x=180^circ##

##rarr2x+5/2*x=180^circ##

##rarr2x+(5x)/2=180^circ##

##rarr(9x)/2=180^circ##

##rarr9x=360^circ##

##color(green)(rArrx=40^circ##

Now we know that the base angles are ##color(green)(40^circ##

So, the other angle is ##color(orange)(2 1/2*40=100)##