Questions and Answers #1 Right Triangles & Trigonometry

Questions and Answers Sheet 1 Right Triangles & Trigonometry Question #1 What is the cot 0 ? Answer: It isn't defined, as tan 0 = 0 and cot 𝜃 = cot 0 = 1 1 tan 𝜃 . 0 Question #2 We have a right triangle. (l ) = 4.5cm 1 (l ) = 4.5 3 2 What are the measures of the angles? Answer: This is the special right triangle because one leg is sqrt3 times longer than the other. Thus, the angles are: 30∘ , 60∘ and 90∘ Question #3 Given a right triangle. anglea is 51∘ . A line is drawn from anglec (90∘ ) to the hypotenuse, creating a 7∘ angle between the line and the cathetus adjusted to angle b. The cathetus adjusted to angel is 47cm. Find the length of the line drawn. Answer: Angle 𝑏 = 180∘ -angle c-angle a angle 𝑏 = 180∘ − 90∘ − 51∘ angle 𝑏 == 46∘ Let the line be x. Use the law of Sines: (sin 4 6∘ )/47 = (sin 5 1∘ )/𝑥 𝑥 = (47 sin 5 1∘ )/ sin 4 6∘ 𝑥 ≈ 50.8𝑐𝑚 Question #4 Given a rectangle ( 17.5𝑐𝑚 × 26.2𝑐𝑚 ). Find the angle between the longer side and the diagonal. Answer: Let the needed angle be B, the length(17.5cm) be l and the width (26.2cm) be w. The diagonal creates a right triangle. Use the formula: tan 𝐵 = 𝑙/𝑤 tan 𝐵 = 17.5/26.2 𝐵 = tan−1 (17.5/26.2) 𝐵 ≈ 33.7∘ Question #5 What is the value of the following expression: √3 cot 1 5∘ Answer: First, find tan 1 5∘ tan(𝐴 − 𝐵) = (tan 𝐴 − tan 𝐵)/(1 + tan 𝐴 tan 𝐵) tan 1 5∘ = tan(60∘ − 45∘ ) tan 1 5∘ = (tan 6 0∘ − tan 4 5∘ )/(1 + tan 6 0∘ tan 4 5∘ ) tan 1 5∘ = (√3 − 1)/ (1 + (√3)(1)) = (√3 − 1)/(√3 + 1) Then, cot 1 5∘ = (√3 + 1)/(√3 − 1) cot 1 5∘ = (√3 + 1)/(√3 − 1) × (√3 + 1)/(√3 + 1) cot 1 5∘ = (3 + 2√3 + 1)/(3 − 1) cot 1 5∘ = (2√3 + 4)/2 cot 15∘ = √3 + 2 Substitute to given expression: √3𝑐𝑜𝑡15∘ = √3(√3 + 2) √3𝑐𝑜𝑡15∘ = 3 + 2√2 Question #6 Given a right triangle. One cathetus is 100cm long. Find the length of the other cathetus, if the angle opposite to it is 71.6∘ . Round your answer to an integer. Answer: Let the unknown cathetus be a. Use the tangent ratio: tan 7 1.6∘ = 𝑎/100 𝑎 = 100 tan 7 1.6∘ 𝑎 ≈ 300 cm Question #7 Given a right triangle with angle 𝑎 = 23.4∘ . The cathetus opposite to anglea is 𝐴̅ = 5.75 m. Find the second cathetus (𝐵̅ ). Answer: Since it's a right triangle, use the tangent ratio: ̅̅̅̅̅) tan(∠) = (𝑜𝑝𝑝 ̅̅̅̅̅)/(𝑎𝑑𝑗 Substitute values from the given: tan 2 3.4∘ = 5.75/𝐵̅ Isolate 𝐵̅ : 𝐵̅ = 5.75/(tan 2 3.4∘ ) 𝐵̅ ≈ 13.29𝑚 Question #8 Given a right triangle with angle 𝑎 = 65∘ . The cathetus opposite to angle a is 𝐴̅ = 250 m. Find the second cathetus (𝐵̅ ). Answer: Since it's a right triangle, use the tangent ratio: ̅̅̅̅̅̅̅̅)/((𝑎𝑑𝑗) ̅̅̅̅̅̅̅) tan(∠) = ((𝑜𝑝𝑝) Substitute values from the given: tan 6 5∘ = 250/𝐵̅ Isolate 𝐵̅ : 𝐵̅ = 250/(tan 6 5∘ ) 𝐵̅ ≈ 116.6 m Question #9 Given a right triangle, where one cathetus is 𝐴̅ = 51 m and the second cathetus is 𝐵̅ = 39 m. Find the hypothenuse (𝐶̅ ) and the angle opposite to 𝐵̅(∠𝑎) Answer: To find 𝐶̅ , use the Pythagorean Theorem: ̅̅̅ 𝐴2 + ̅𝐵̅̅2̅ = ̅𝐶̅̅2̅ ̅𝐶̅̅2̅ = 512 + 392 𝐶̅ = √4122 𝐶̅ = 64.2 m To find angle a, use the tangent ratio: tan ∠𝑎 = 𝐵̅ /𝐴̅ ∠𝑎 = 𝑡𝑎𝑛(−1) (𝐵̅ /𝐴̅) ∠𝑎 = 𝑡𝑎𝑛(−1) (13/17) ∠𝑎 ≈ 37.4∘ Question #10 Given a right triangle, where one cathetus is 𝐴̅ = 11 m, the hypothenuse is 𝐶̅ = 15 m, and ∠𝑐 = 90∘ . Find all missing sides (𝐵̅ ) and angles (angle a and angle b). Answer: first, to find 𝐵̅ , use the Pythagorean Theorem: ̅̅̅ 𝐴2 + ̅𝐵̅̅2̅ = ̅𝐶̅̅2̅ Isolate ̅𝐵̅̅2̅ ̅𝐵̅̅2̅ = ̅𝐶̅̅2̅ − ̅̅̅ 𝐴2 ̅𝐵̅̅2̅ = 152 − 112 𝐵̅ = √104 𝐵̅ ≈ 10.2𝑚 To find angle a, using the cosine ratio (assuming 𝐴̅ is adj to angle a ): cos angle 𝑎 = 𝐴̅/𝐶̅ = 11/15 angle a=cos(-1)(11/15) ≈ 43∘ To find angle b, subtract both other angles from 180∘ (since the sum of all angles in any triangle is 180∘ ): ∠𝑏 = 180∘ − ∠𝑎 − ∠𝑐 ∠𝑏 = 180∘ − 43∘ − 90∘ ∠𝑏 = 47∘ Answer: 𝐵̅ ≈ 10.2𝑚; ∠𝑎 ≈ 43∘ ; ∠𝑏 ≈ 47∘