Reasoning · Grade 4-1 Polygons and Angles

Problem

Chase angles through triangles in a crossing figure

Points A, E and D sit in order on one line across the top, with B below left and C below right. Segments AC, EB, EC and DB are drawn; F is where AC meets EB, G where AC meets DB, and H where EC meets DB. The given angles are ∠EAC = 50°, ∠EDB = 30°, ∠EBD = 24° and ∠ACE = 25°. Find ∠BFG and then ∠EHD.
50° 30° 24° 25° A E D B C F G H
Your answer
How to solve
Strategy Identify Subproblems — The tangle of five segments hides several ordinary triangles. The trick is to hunt for a triangle that already contains two of the given angles, use the 180° sum to fill in its third angle, and then hand that new angle on to a neighbouring triangle that touches the angle I actually want. Each unknown needs its own two-triangle chain, so I solve ∠BFG and ∠EHD as two separate small problems rather than one big one.
1STEP 1

Use the top line to relabel the given angles

The top line lets each given angle be renamed as an equal one.

∠ DAC = ∠ EAC = 50°, ∠ ADB = ∠ EDB = 30°
2STEP 2

Find the angle at G from triangle AGD

In the big triangle the crossing angle is 100 degrees.

∠ AGD = 180° - 50° - 30° = 100°
3STEP 3

Swing that 100° across to the other side of DB

Swinging to the other side gives 80 degrees.

∠ AGB = 180° - 100° = 80°
4STEP 4

Finish ∠BFG in triangle FGB

The next triangle gives the first answer, 76 degrees.

∠ BFG = 180° - 80° - 24° = 76°
5STEP 5

Start the second chase in triangle AEC

The second chase starts from 105 degrees.

∠ AEC = 180° - 50° - 25° = 105°
6STEP 6

Turn 105° into the angle on the other side of E

Its neighbour is 75 degrees.

∠ DEC = 180° - 105° = 75°
7STEP 7

Finish ∠EHD in triangle EHD

The last triangle makes the second answer 75 degrees.

∠ EHD = 180° - 75° - 30° = 75°
Answer
76, 75 degrees
180 − 80 − 24 = 76, 180 − 75 − 30 = 75
Both answers are between 0° and 180° and both are acute, matching the fairly narrow arcs drawn at F and H in the figure. A second check on ∠BFG: F is a crossing point, so the angle vertically opposite it, ∠AFE, must also be 76°; then triangle AFE gives ∠AEF = 180° - 50° - 76° = 54°, and ∠AEB = 54° is plausible for the steep segment EB in the picture. A second check on ∠EHD: its neighbour ∠EHB on the straight segment DB is 180° - 75° = 105°, and in triangle EHB that leaves ∠HEB = 180° - 105° - 24° = 51°, safely positive — so no angle in the chain comes out impossible.
Takeaway

In a crowded figure, hunt for the one triangle that already holds two known angles — then pass its answer along to the next triangle like a relay baton.

  • Use the top line to relabel the given angles
  • Find the angle at G from triangle AGD
  • Swing that 100° across to the other side of DB
  • Finish ∠BFG in triangle FGB
  • Start the second chase in triangle AEC
  • Turn 105° into the angle on the other side of E
  • Finish ∠EHD in triangle EHD