University:
High School
Course:
High School Mathematics
Academic year:

2023
Views:

55

Pages:

3
Author:

anon

Quiz: Angle Relationships in Geometric Figures 1. Find the missing values. To earn full marks, calculations must be shown and reasons given for each answer. (½ mark calculation/answer, ½ mark justification). a) b) c) Sum of ext. Angles is ◦ 360 ? ◦ ◦ ◦ (◦ ◦ 𝑥 = 360 − 165 + 155 𝑐 = 75 𝑎=𝑐 ◦ ) ◦ ◦ = 40 ◦ 𝑏 = 75 d) e) Opposite angles are equal ◦ ∴𝑎 = 55 ◦ ◦ ◦ 55 + 55 + 𝑏 + 𝑏 = 360 (𝑏 = 𝑐) ◦ ◦ → 110 + 2𝑏 = 360 ◦ ◦ 2𝑏 = 360 − 110 ◦ 𝑦 = 44 (is osseles triangle) ◦ ◦ 𝑤 = 180 − 88 ◦ = 92 𝑥=𝑤+𝑦 ◦ ◦ = 92 + 44 ◦ = 136 ◦ ∴𝑏 = 125 𝑎𝑛𝑑𝑐 = 125 ◦ f) ◦ 2𝑏 = 250 ◦ = 134 An exterior angle in a triangle is equal to the sun of oppo. Interior angles = 360 − 320 = 75 ◦ 𝑦 = 61 + 73 ◦ 1. a and 119 are sup. Angles. ◦ ◦ 𝑎 = 180 − 119 ◦ ∴𝑎 = 61 2. ◦ ◦ ◦ ◦ ◦ 61 + 𝑐 + 191 = 360 ◦ ◦ ◦ ◦ 𝑐 + 252 = 360 𝑐 = 360 − 252 ◦ ∴𝑐 = 108 2. Determine the values of each variable. Justify your answers. ◦ 1. 𝑥 + 3𝑥 − 22 = 180 ◦ ◦ 4𝑥 = 180 + 22 ◦ 𝑋= 202 4 ◦ ∴𝑥 = 50. 5 ◦ 𝑎 + 𝑐 + 87 + 104 = 360 ◦ ◦ 2. 3𝑥 − 22 = 151. 5 − 22 ◦ ◦ = 129. 5 ◦ ◦ ◦ 2𝑥 − 10 = 101 − 10 ◦ = 91 ◦ ◦ 𝑥 + 15 = 65. 5 ◦ ◦ 3. ∴𝑦 = 180 − 91 ◦ = 89 ◦ ◦ ∴𝑧 = 180 − 65. 5 ◦ = 114. 5 ◦ ◦ ( ◦ 4. ∴𝑤 = 360 − 129. 5 + 91 + 65. 5 ◦ ◦ ) ◦ = 360 − 286 ◦ = 74 ◦ ◦ ∴𝑢 = 180 − 74 ◦ = 106 3. Explain why a triangle cannot have two obtuse interior angles. ◦ Because the sum of two interior angles will be greater than 180 4. Find the sum of the interior angles of a polygon with 15 sides. n=15 ◦ ◦ ∴(𝑛 − 2) × 180 (𝑛 − 2) × 180 ◦ = (15 − 2) × 180 ◦ = 13 × 180 ◦ = 2340 5. Find the measure of each interior angle of a regular polygon with 12 sides. 𝑛 = 12 ◦ = (12 − 2) × 180 ◦ = 10 × 180 ◦ = 1800 ◦ ∴𝐼𝑛𝑡𝑒𝑟𝑖𝑜𝑟 𝑎𝑛𝑔𝑙𝑒 = ◦ = 150 Bonus (to be completed ONLY after all other questions have been answered). Calculate the sum of ∠ABC and ∠ADC. Justify your answer. 1800 12 ◦ 𝑥 + 𝑥 + 𝑡 = 180 ◦ 2𝑥 + 𝑡 = 180 ◦ 𝑡 = 180 − 2𝑥 t and z are supplementary angles. ◦ So, 𝑡 + 𝑧 = 180 ◦ ◦ ◦ 180 − 2𝑥 + 180 − 2𝑦 = 180 ◦ ◦ ◦ ◦ − 2𝑥 − 2𝑦 + 360 = 180 − 2𝑥 − 2𝑦 = 180 − 360 ◦ − 2𝑥 − 2𝑦 =− 180 −2𝑥 2 − −2𝑦 2 ◦ ◦ = −180 2 𝑦 + 𝑦 + 𝑧 = 180 ◦ 2𝑦 + 𝑧 = 180 ◦ ∴𝑥 + 𝑦 = 90 ◦ ∠𝐴𝐵𝐶 + ∠𝐴𝐷𝐶 = 90 ◦ 𝑧 = 180 − 2𝑦