Maths Olympiad Prep

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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.

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