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: 48. From a point T, the bearings of two cities M and N are N67°W and N23°E respectively. If the bearing of N from M is N70°E and |TM| = 160m, calculate |TN| to the nearest metre. Step 1: Determine the angle MTN. The bearing of M from T is N67°W, meaning M is 67^ West of North from T. The bearing of N from T is N23°E, meaning N is 23^ East of North from T. The angle between these two bearings is MTN = 67^ + 23^ = 90^. Thus, MTN is a right-angled triangle at T. Step 2: Determine the angle TMN. The bearing of M from T is N67°W. This means the bearing of T from M is S67°E. From M, draw a North line. The line MT makes an angle of 67^ with the South line (towards East). This means the angle from the North line at M, measured clockwise to MT, is 180^ - 67^ = 113^. The bearing of N from M is N70°E. This means the angle from the North line at M, measured clockwise to MN, is 70^. Since both angles are measured clockwise from the North line at M and are in the eastern direction, the angle TMN is the difference between these two angles: TMN = 113^ - 70^ = 43^ Step 3: Calculate |TN| using trigonometry in the right-angled triangle MTN. We have |TM| = 160m, MTN = 90^, and TMN = 43^. ( TMN) = oppositeadjacent = (|TN|)/(|TM|) (43^) = (|TN|)/(160) |TN| = 160 × (43^) |TN| ≈ 160 × 0.9325 |TN| ≈ 149.2 m Step 4: Round to the nearest metre. |TN| ≈ 149 m Comparing this to the options, 140m (D) is the closest. There might be a slight discrepancy in the problem's values or options. We will select the closest option. The value of |TN| is approximately 149 m. 49. Use the information below to answer the question. The roots of the equation 2x^2 - 10x + 8 = 0 are and . Calculate (1)/() + (1)/(). Step 1: Simplify the expression (1)/() + (1)/(). (1)/() + (1)/() = ( + )/() Step 2: Identify the coefficients of the quadratic equation ax^2 + bx + c = 0. For 2x^2 - 10x + 8 = 0, we have a=2, b=-10, c=8. Step 3: Calculate the sum of the roots ( + ) and the product of the roots (). + = -(b)/(a) = -(-10)/(2) = 5 = (c)/(a) = (8)/(2) = 4 Step 4: Substitute the values into the simplified expression. ( + )/() = (5)/(4) The correct option is A. The value is (5)/(4). 50. A man is four times as old as his son. If the difference between their ages is 36, find the sum of their ages. Step 1: Define variables for their ages. Let the son's age be s. Let the man's age be m. Step 2: Formulate equations based on the given information. "A man is four times as old as his son": m = 4s "The difference between their ages is 36": m - s = 36 Step 3: Solve the system of equations. Substitute the first equation into the second: 4s - s = 36 3s = 36 s = (36)/(3) s = 12 Step 4: Find the man's age. m = 4s = 4 × 12 = 48 Step 5: Calculate the sum of their ages. Sum = m + s = 48 + 12 = 60 The correct option is B. The sum of their ages is 60. 51. Two cities are both on longitude 15^ E. If their latitudes are 19^ N and 44^ S and R = 6,400 km, find their distance apart in km. Give your answer in terms of . Step 1: Determine the angular difference in latitude. Since the cities are in different hemispheres (North and South), add their latitudes. = 19^ + 44^ = 63^ Step 2: Convert the angular difference from degrees to radians. _rad = 63 × ()/(180) Step 3: Calculate the distance using the formula D = R × _rad, where R is the radius of the Earth. D = 6400 km × 63 × ()/(180) D = (6400 × 63)/(180) D = (403200)/(180) D = 2240 km The correct option is A. The distance apart is 2240 km. 52. If = (12)/(5) and is acute, evaluate ( - )/( + ). Step 1: Divide the numerator and denominator of the expression by . ( - )/( + ) = ( )/( ) - ( )/( )( )/( ) + ( )/( ) = ( - 1)/( + 1) Step 2: Substitute the given value of = (12)/(5) into the simplified expression. = (12)/(5) - 1(12)/(5) + 1 Step 3: Simplify the expression. = (12 - 5)/(5)(12 + 5)/(5) = (7)/(5)(17)/(5) = (7)/(17) The correct option is B. The value is (7)/(17). 53. Find the equation of a curve which passes through the point (2, -3) and its gradient at any point is given by (dy)/(dx) = 3x^2 - 6x - 5. Step 1: Integrate the gradient function to find the equation of the curve y. y = (3x^2 - 6x - 5)\,dx y = 3x^2+12+1 - 6x^1+11+1 - 5x + C y = x^3 - 3x^2 - 5x + C Step 2: Use the given point (2, -3) to find the constant of integration C. Substitute x=2 and y=-3 into the equation: -3 = (2)^3 - 3(2)^2 - 5(2) + C -3 = 8 - 3(4) - 10 + C -3 = 8 - 12 - 10 + C -3 = -14 + C Step 3: Solve for C. C = -3 + 14 C = 11 Step 4: Write the final equation of the curve. y = x^3 - 3x^2 - 5x + 11 The correct option is E. The equation of the curve is y = x^3 - 3x^2 - 5x + 11. 54. The angle of a sector of a circle with radius 9.1 cm is 100^. Find the perimeter of the sector, correct to the nearest centimetre. (Take = (22)/(7)) Step 1: Convert the angle from degrees to radians. _rad = 100^ × ()/(180^) = (100)/(180) = (5)/(9) Step 2: Calculate the arc length (L) of the sector using the formula L = r_rad. L = 9.1 cm × (5)/(9) Substitute = (22)/(7): L = 9.1 × (5)/(9) × (22)/(7) L = (91)/(10) × (5)/(9) × (22)/(7) L = (13)/(1) × (1)/(9) × (11)/(1) = (143)/(9) cm L ≈ 15.889 cm Step 3: Calculate the perimeter (P) of the sector. The perimeter is the sum of the arc length and two radii. P = L + 2r P = (143)/(9) + 2(9.1) P = 15.889 + 18.2 P = 34.089 cm Step 4: Round the perimeter to the nearest centimetre. P ≈ 34 cm The correct option is D. The perimeter of the sector is 34 cm. Drop the next question.