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Quiz #5 08/12/
Name:
Directions: Make sure to read each problem carefully. To receive full credit, you must show all your work.
Problem 1. Complete the triangle with side lengths a = 8, b = 9, c = 6. (To reiterate, we usually make angle A opposite of side a, etc.)
Using the law of cosines,
82 = 9^2 +6^2 −2(9)(6) cos A ⇒ −53 = −108 cos A ⇒ cos A =. 49074 ⇒ A = cos−^1. 49074 ≈ 60. 61 o
92 = 8^2 +6^2 −2(8)(6) cos B ⇒ −19 = −96 cos A ⇒ cos B =. 19792 ⇒ B = cos−^1. 19792 ≈ 78. 58 o
⇒ C ≈ 40. 81 o
Problem 2. Given: AE = CE and AB = CB, Prove: ∠ADB = ∠CDB.
Statement Reason AE = CE given AB = CB given BE = BE anything equals itself ∆AEB ∼= ∆CEB SSS ∠AEB = ∠CEB because the triangles are congruent ∠AEB = ∠CEB = 90o^ because AC is a straight line, and they must add to 180o ∠AED = 90o^ because the sum of ∠AED and ∠AEB is 180 o ∠CED = 90o^ because the sum of ∠CED and ∠CEB is 180 o ED = ED anything equals itself ∆AED = ∆CED SAS ∠ADE = ∠CDE similar triangles ∠ADB = ∠ADE and ∠CDB = ∠CDE the angles created are the same ∠ADB = ∠CDB ∠ADB = ∠ADE = ∠CDE = ∠CDB So they weren’t all that easy. There are some unnecessary steps here but, you get the idea.
