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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Question 1 1.1 The ratio in a right-angled triangle is defined as the length of the side opposite to the angle divided by the length of the hypotenuse. = OppositeHypotenuse 1.2 Step 1: Identify the sides relative to the 60^ angle in the given special triangle. For the 60^ angle: • The side opposite is sqrt(3). • The side adjacent is 1. Step 2: Use the definition of . = OppositeAdjacent 60^ = sqrt(3)1 60^ = sqrt(3) 1.3 Step 1: Use the given information to determine the quadrant and coordinates. Given = (12)/(5) and 180^ < < 360^. Since is positive, must be in Quadrant III (where both x and y are negative). We know = (x)/(y). So, we can set x = -12 and y = -5. Step 2: Calculate the hypotenuse r. Using the Pythagorean theorem r^2 = x^2 + y^2: r^2 = (-12)^2 + (-5)^2 r^2 = 144 + 25 r^2 = 169 r = sqrt(169) r = 13 Step 3: Determine the value of . = (x)/(r) = (-12)/(13) = -(12)/(13) Question 2 2.1 Step 1: Isolate (x+10^). 2 (x+10^) = 1.532 (x+10^) = (1.532)/(2) (x+10^) = 0.766 Step 2: Find the reference angle. Let = x+10^. = (0.766) ≈ 49.99^ Step 3: Find x within the given range 0^ < x < 90^. Since 0^ < x < 90^, then 10^ < x+10^ < 100^. Since (x+10^) is positive, x+10^ can be in Quadrant I or Quadrant II. For Quadrant I: x+10^ = 49.99^ x = 49.99^ - 10^ x = 39.99^ Rounding to one decimal place: x = 40.0^ (The Quadrant II solution x+10^ = 180^ - 49.99^ = 130.01^ x = 120.01^ is outside the given range for x.) 2.2 Step 1: Recall the values of special angles. 45^ = sqrt(2)2 45^ = sqrt(2)2 60^ = sqrt(3) 30^ = (1)/( 30^) = (1)/(1/sqrt(3)) = sqrt(3) Step 2: Substitute the values into the expression. ( 45^ · 45^)/( 60^ · 30^) = (sqrt(2)2) · (sqrt(2)2)sqrt(3) · sqrt(3) Step 3: Simplify the expression. = (2)/(4)3 = (1)/(2)3 = (1)/(2) × (1)/(3) = (1)/(6) Question 3 3.1 A neat, labeled diagram representing the scenario: (A) at (0,0); (B) at (5,0); (C) at (5,3.5); (A) node[below] A -- (B) node[below] B -- (C) node[right] C -- cycle; (B) rectangle +(0.2,0.2); (A) -- (C); [draw, angle radius=1cm, "35^"] angle = C--A--B; at (2.5,-0.3) 50 m; at (5.3,1.75) BC; (Diagram shows a right-angled triangle ABC, with angle B = 90 degrees, AB = 50m, angle A = 35 degrees, and BC as the height of the tower.) 3.2 Step 1: Identify the relevant trigonometric ratio. In the right-angled triangle ABC, we have the adjacent side AB = 50 m and the angle A = 35^. We need to find the opposite side BC. We use the tangent function: A = OppositeAdjacent. 35^ = (BC)/(AB) Step 2: Substitute values and solve for BC. 35^ = (BC)/(50 m) BC = 50 m × 35^ BC ≈ 50 m × 0.7002075 BC ≈ 35.010375 m Rounding to one decimal place: BC = 35.0 m 3.3 Step 1: Set up the new scenario with point D. Assume point