2018-I-08 Posted on 16-06-2021 By app.cch No Comments on 2018-I-08 Ans: x=180∘–θ, y=2θ–180∘Consider cyclic quadrilateral ABCD. x+θ=180∘(opp. ∠s, cyclic quad.)x=180∘–θ Since AB//ED, we have ∠ADE=x(alt. ∠s, AB//ED)∠ADE=180∘–θ Consider cyclic quadrilateral BCD. ∠BED+θ=180∘(opp. ∠s, cyclic quad.)∠BED=180∘–θ In ΔDEF, y=180∘–∠FED–∠FDE(∠ sum of Δ)y=180∘–(180∘−θ)–(180∘–θ)y=2θ–180∘ Same Topic: 2018-II-22 2018-II-39 2020-I-18 2021-II-39 2018, HKDSE-MATH, Paper 1 Tags:Properties of Circles