This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Here are the solutions for the triangles: For both triangles, we will assume that the angles that appear to be right angles in the diagrams are indeed 90^. a) Find BC Given a triangle ABC with side AB = 2 and C = 45^. We assume A = 90^. Step 1: Identify the known values and the unknown value. We have a right-angled triangle with the side opposite to C (AB = 2) and the angle C = 45^. We need to find the hypotenuse BC. Step 2: Apply the sine trigonometric ratio. The sine of an angle in a right-angled triangle is the ratio of the length of the opposite side to the length of the hypotenuse. C = oppositehypotenuse = (AB)/(BC) Substitute the given values: 45^ = (2)/(BC) Step 3: Solve for BC. We know that 45^ = sqrt(2)2. sqrt(2)2 = (2)/(BC) Multiply both sides by 2 · BC: BC · sqrt(2) = 2 · 2 BC · sqrt(2) = 4 Divide by sqrt(2): BC = (4)/(sqrt(2)) Rationalize the denominator by multiplying the numerator and denominator by sqrt(2): BC = (4)/(sqrt(2)) · sqrt(2)sqrt(2) = 4sqrt(2)2 BC = 2sqrt(2) The value of BC is 2sqrt(2). b) Find a and c Given a triangle PQR with side PR = 7 and Q = 75^. Side PQ = a and side QR = c. We assume P = 90^. Step 1: Identify the known values and the unknown values. We have a right-angled triangle with the side opposite to Q (PR = 7) and the angle Q = 75^. We need to find the adjacent side PQ = a and the hypotenuse QR = c. Step 2: Find the third angle, R. The sum of angles in a triangle is 180^. R = 180^ - P - Q R = 180^ - 90^ - 75^ R = 15^ Step 3: Find 'a' (side PQ) using the tangent trigonometric ratio. For Q, PR is the opposite side and PQ is the adjacent side. Q = oppositeadjacent = (PR)/(PQ) Substitute the given values: 75^ = (7)/(a) a = (7)/( 75^) We know that 75^ = 2 + sqrt(3). a = (7)/(2 + sqrt(3)) Rationalize the denominator: a = (7)/(2 + sqrt(3)) · 2 - sqrt(3)2 - sqrt(3) = 7(2 - sqrt(3))2^2 - (sqrt(3))^2 = 7(2 - sqrt(3))4 - 3 a = 7(2 - sqrt(3)) Step 4: Find 'c' (side QR) using the sine trigonometric ratio. For Q, PR is the opposite side and QR is the hypotenuse. Q = oppositehypotenuse = (PR)/(QR) Substitute the given values: 75^ = (7)/(c) c = (7)/( 75^) We know that 75^ = sqrt(6) + sqrt(2)4. c = (7)/(sqrt(6) + 2)4 = (28)/(sqrt(6) + 2) Rationalize the denominator: c = (28)/(sqrt(6) + 2) · sqrt(6) - sqrt(2)sqrt(6) - sqrt(2) = 28(sqrt(6) - sqrt(2))(sqrt(6))^2 - (sqrt(2))^2 = 28(sqrt(6) - sqrt(2))6 - 2 c = 28(sqrt(6) - sqrt(2))4 c = 7(sqrt(6) - sqrt(2)) The values are a = 7(2 - sqrt(3)) and c = 7(sqrt(6) - sqrt(2)). Send me the next one 📸