Maths Olympiad Prep

Library / Macedonian Mathematical Competitions /25 of 73

Macedonian Mathematical Competitions

Geometry Difficulty 4.7 AIME Prove it North Macedonia

Calculate the angles of the triangle if it is known that the sum of two of its angles is 56\frac{5}{6} from the right angle and one of these angles is 20° bigger than the other.

Solution

We have that α+β=56⋅90∘=75∘\alpha + \beta = \frac{5}{6} \cdot 90^\circ = 75^\circ and α−β=20∘\alpha - \beta = 20^\circ.

Then β=(75∘−20∘):2=27∘30′\beta = (75^\circ - 20^\circ) : 2 = 27^\circ 30' and α=75∘−27∘30′=47∘30′\alpha = 75^\circ - 27^\circ 30' = 47^\circ 30', so we have that γ=180∘−(α+β)=180∘−75∘=105∘\gamma = 180^\circ - (\alpha + \beta) = 180^\circ - 75^\circ = 105^\circ.

The triangle is obtuse-angled.

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.