Reasoning · Grade 4-2 Parallel Lines and Angles

Problem

Chase angles using a rectangle's parallel sides

Two straight segments drawn inside rectangle ABCD cross each other. The first slants from the top side down to the bottom side, making 130 degrees with the top side on the A side. The second rises from the bottom side to vertex D, making 150 degrees with the bottom side on the B side. Find the angle where they cross, then the angle at D between that segment and side DC.
A D B C 130° 150° 1 2
Your answer
How to solve
Strategy Draw a Diagram — First I redraw the figure and give the three unnamed points names — P where the first segment leaves the top side, Q where it lands on the bottom side, R where the second segment leaves the bottom side, and X where the two segments cross. With names in place the tangle of lines splits into two easy subproblems: the right triangle RCD, which alone settles ②, and the small triangle XQR sitting on the bottom side, which settles ①. Each triangle needs nothing more than the 180° angle sum plus one straight-line or parallel-line fact to bring the printed measure to where it is wanted.
1STEP 1

Name the four working points

Give the four working points names.

∠ APQ = 130°, ∠ BRD = 150°
2STEP 2

Turn the 150° around at R

Turning the 150 gives a neighbour of 30 degrees.

∠ DRC = 180° - 150° = 30°
3STEP 3

Solve the right triangle RCD to get ②

In the right triangle the second angle is 60 degrees.

② = ∠ RDC = 180° - 90° - 30° = 60°
4STEP 4

Carry the 130° from the top side down to the bottom side

The parallel sides let the 130 be carried down.

∠ PQC = ∠ APQ = 130°
5STEP 5

Turn that 130° around at Q

Turning it around gives 50 degrees.

∠ PQR = ∠ PQB = 180° - 130° = 50°
6STEP 6

Solve the small triangle XQR

In the small triangle the crossing angle is 100 degrees.

∠ RXQ = 180° - 30° - 50° = 100°
7STEP 7

Flip at the crossing point to get ①

Flipping to the other side gives 80 degrees.

① = 180° - ∠ RXQ = 180° - 100° = 80°
Answer
80, 60 degrees
180 − 100 = 80, 180 − 90 − 30 = 60
Both answers are degree measures between 0° and 180°, as any angle inside a figure must be. They match the picture: the arc at ② looks a little wider than half a right angle, and 60° is exactly that; the arc at ① looks a bit less than a right angle, and 80° is exactly that. The whole chain also stays consistent — in triangle XQR the angles 30° + 50° + 100° = 180°, and in triangle RCD 30° + 90° + 60° = 180°. There is a neat cross-check too: ① is the exterior angle of triangle XQR at X, so it should equal the sum of the two far corners, 30° + 50° = 80°, which is what came out. Finally, notice that ② never used the 130° at all — it depends only on the 150°, which is right, because moving the first segment does not move D or R.
Takeaway

Parallel sides carry an angle across the figure for free, and every triangle you find hands you its last corner for 180° minus the other two.

  • Name the four working points
  • Turn the 150° around at R
  • Solve the right triangle RCD to get ②
  • Carry the 130° from the top side down to the bottom side
  • Turn that 130° around at Q
  • Solve the small triangle XQR
  • Flip at the crossing point to get ①