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 to the problems. 1.1.1 Calculate the value of x. Step 1: Use the distance formula from the origin O(0;0) to point P(x; sqrt(3)). The distance OP is given as 2. OP = sqrt((x_P - x_O)^2 + (y_P - y_O)^2) 2 = sqrt((x - 0)^2 + (3) - 0)^2 2 = sqrt(x^2 + (3))^2 Step 2: Simplify and solve for x. 2 = sqrt(x^2 + 3) Square both sides: 2^2 = x^2 + 3 4 = x^2 + 3 x^2 = 4 - 3 x^2 = 1 x = ± 1 Step 3: From the diagram, point P is in the first quadrant, where the x-coordinate is positive. Therefore, x = 1. The value of x is 1. 1.1.2 Hence, calculate the size of . Step 1: First, determine the angle for point P(1; sqrt(3)). Let this angle be _P. We know OP = 2. _P = (x)/(OP) = (1)/(2) _P = (y)/(OP) = sqrt(3)2 This means _P = 60^. Step 2: We are given that POQ = 90^. From the diagram, point Q is in the fourth quadrant, which means the angle for Q is obtained by rotating P clockwise by 90^. Let _Q be the angle for point Q. _Q = _P - 90^ _Q = 60^ - 90^ _Q = -30^ Step 3: The angle is defined as TOQ, which is the acute angle between the positive x-axis (line OT) and OQ. Since _Q = -30^, this means OQ is 30^ clockwise from the positive x-axis. Therefore, = 30^. The size of is 30^. 1.1.3 Determine the coordinates of Q. Step 1: Point Q has coordinates (a; b). We know OQ = 20 and the angle for Q is _Q = -30^. The coordinates (a; b) can be found using: a = OQ _Q b = OQ _Q Step 2: Substitute the values. a = 20 (-30^) b = 20 (-30^) Step 3: Evaluate the cosine and sine values. (-30^) = (30^) = sqrt(3)2 (-30^) = -(30^) = -(1)/(2) Step 4: Calculate a and b. a = 20 (sqrt(3)2) = 10sqrt(3) b = 20 (-(1)/(2)) = -10 The coordinates of Q are (10sqrt(3); -10). That's 2 down. 3 left today — send the next one.