Problem
At C, the square gives angle BCD = 90° and the triangle gives angle DCE = 60°, so angle BCE = 90 + 60 = 150 degrees.
Angles that meet at one point add up, so the square angle plus the triangle angle gives the whole angle at C.
4.MD.C.7Identify SubproblemsBC = CE (square side = triangle side), so triangle BCE is isosceles: angle CBE = angle CEB = (180 - 150) / 2 = 15 degrees.
Equal sides give equal base angles, and the three angles still total 180 degrees.
4.MD.C.7Identify SubproblemsThe two base angles of triangle BCE, angle CBE and angle CEB, are each 15 degrees.
Triangle BCE is isosceles: sides BC and CE have the same length, and equal sides sit opposite equal angles, so the two base angles must match.
BC equals CE because BC equals the square's side DC, and DC equals the triangle's side CE, so the two outer sides match through the shared side.
When two sides of a triangle are equal you can fold the triangle along the line down its middle, and one equal side lands exactly on the other, carrying its base angle onto the other base angle.
The angle at C is 150 degrees, and the three angles of the triangle add to 180, so the two equal base angles share the 30 degrees that are left over, giving 15 each.
The whole angle BCE at corner C is 150 degrees, made of the square's 90-degree corner and the triangle's 60-degree corner placed right next to each other with no gap.
Two angles meeting at the same point with no gap and no overlap join into one larger angle equal to their sum.
The three angles inside triangle BCE must total 180 degrees, so once the top angle is fixed the rest is decided.
Triangle BFC: angle FCB = 90°, angle FBC = 15°, so angle BFC = 75°; angle a is its straight-line partner: 180 - 75 = 105 degrees.
Once two angles of triangle BFC are known the third follows, and angle a is its straight-line partner along DC.
4.MD.C.7Identify SubproblemsThis only needs Grade 4 angle-adding plus spotting that a square side and a triangle side are equal!