Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3135859

tan60 = a/(0.5*8) a = 4tan60 a = 4√(3) cm. a is the diameter of the circle. b = ½(a) b = 2√(3) cm. b is the radius of the circle. c = ¼(8) c = 2 cm. c is the radius of the two equal half circles. Calculating shaded area. It is; 2(area sector with radius 2√(3) cm and angle 120° - area triangle with height 2√(3) cm and base 2√(3)sin120 cm)+2(area sector with radius 2√(3) cm and angle 60° - area triangle with height 2√(3) cm and base 2√(3)sin60)+2(area sector with radius 2 cm and angle 60° -...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3135856

a = 4+4 a = 8 units. a is the radius of the quarter circle. cosb = ⅛(0.5*4) b = acos(¼)° cosb = c/8 c = 8cos(acos(¼)) c = ¼(8) c = 2 units. c is the width of the inscribed rectangle. sinb = d/8 d = 8sin(acos(¼)) d = 7.74596669241 units. Therefore, area inscribed rectangle is; cd = 2*7.74596669241 = 15.4919333848 square units.r

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3135177

Let AD = BC = 1 unit. a = 180-100-30 a = 50° a is angle BAC. (1/sin50) = (b/sin30) b = 0.65270364467 units. b is AC. Therefore, the required angle theta is, let it be c. (0.65270364467/sinc) = (1/sin100) sinc = 0.65270364467sin100 sinc = 0.64278760969 c = asin(0.64278760969) c = 40°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3135172

Let BD = CD = 1 unit. a = 180-45-30 a = 105° a is angle CAD. (1/sin105) = (b/sin45) b = 0.73205080757 units. b is AC. c² = 0.73205080757²+2²-2*2*0.73205080757cos30 c = 1.41421356237 units. c is AB. ( 1.41421356237/sin30) = (2/sind) d = 45° e = 180-d e = 180-45 e = 135° e is angle BAC. Therefore, the required angle alpha is, let it be f. f = e-a f = 135-105 f = 30°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3135154

a = 180-60 a = 120° b = 180-120-45 b = 15° (1/sin15) = (c/sin45) c = 2.73205080757 units. sin60 = d/c d = 2.73205080757sin60 d = 2.36602540378 units. cos60 = e/c e = 2.73205080757cos60 e = 1.36602540378 units.ñ f = 2-e f = 2-1.36602540378 f = 0.63397459622 units. Therefore, the required angle is, let it be g. tang = d/f g = atan(2.36602540378/0.63397459622) g = 75°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3134184

Area blue is; 2(area quarter circle with radius 8 cm - area isosceles right-angled triangle with equal lengths 8 cm) + 2(area quarter circle with radius 4 cm - area isosceles right-angled triangle with equal lengths 4 cm) = 2((¼*8*8π)-(½*8*8))+2((¼*4*4π)-½*4*4)) = 2(16π-32)+2(4π-8) = 2(16π-32+4π-8) = 2(20π-40) = 40(π-2) cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3134182

Notice. 10 m is the radius of the circle. a = 10-3 a = 7 m. sinb = (7/10) b = asin(7/10)° Calculating length AB. tanb = 10/(AB) AB = 10/(tanb) AB = 10/tan(asin(7/10)) AB = 10.2020406122 m. Or 7 ~ 3 10 ~ x Cross Multiply. 7x = 30 x = ⅐(30) m. x is length Be. It implies; y = 10+x y = 10+⅐(30) y = ⅐(100) m. y is length Bc Therefore; z² +10² = (⅐(100))² z² = (10000/49)-100 z² = (10000-4900)/49 z = √(5100/49) z = ⅐(10√(51)) m. z = 10.2020406122 m. z is the required length AB.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3134177

Calculating angle x. Let the longest ascribed quadrilateral side length be 1 unit. a = = 180-25-2(35) a = 180-95 a= 85° (1/sin85) = (b/sin25) b = 0.42423259484 units. c = 180-35-2(25) c = 180-85 c = 95° (1/sin95) = (d/sin50) d = 0.76897060844 units. e² = 0.76897060844²+0.42423259484²-2*0.42423259484*0.76897060844cos35 e = 0.48665963987 units. It implies; (0.48665963987/sin35) = (0.76897060844/sinf) f = asin(0.90630778704) f = 65° g = 180-f g = 180-65 g = 115° Therefore, the required...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3133540

Calculating Area Triangle A. Let the two equal inscribed angles be a each. (0.5*14*12sina)÷(0.5*12*35sina) = 14/35 = 2/5 = 2:5 Where 2:5 is the ratio of the bases of the both triangles with equal angle opposite their base. Let the total length be x. It implies; (2x)² = 14²+12²-2*14*12cosa 4x² = 340-336cosa cosa = (340-4x²)/336 --- (1). (5x)² = 12²+35²-2*12*35cosa 25x² = 1369-840cosa cosa = (1369-25x²)/840 --- (2). Calculating x. Equating (1) and (2). (340-4...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3133538

b = √(a) b = √(48) b = √(16*3) b = 4√(3) units. b is the radius of the ascribed quarter circle. c = ½(b) c = 2√(3) units. c is the radius of the inscribed half circle. tan30 = d/c ⅓√(3) = d/(2√(3)) d = 2 units. e = b-d e = (4√(3)-2) units. e = 4.92820323028 units. e is the side length of the inscribed regular triangle. f = ½(e) f = (2√(3)-1) units. tan60 = g/f g = √(3)(2√(3)-1) g = (6-√(3)) units. g = 4.26794919243 units. g shares the inscribed regular triangle to two equal right-angled t...

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