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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2680115

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Let a be the radius of the circle. tanb = (0.5a)/a b = atan(0.5)° cos(atan(0.5)) = c/6 c = 5.366563146 units. sin(atan(0.5)) = d/6 d = 2.683281573 units. cosatan(0.5) = 3/a a = 3.3541019662 units. Again, a is the radius of the circle. It implies; Area of the rectangle is; ac = 3.3541019662*5.366563146 = 18 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2585239

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a = ⅓(90) a = 30° sin30 = b/4 b = 2 cm. sin60 = c/4 c = 2√(3) cm. d = √(4²-2²)-√(4²-(2√(3))²) d = √(12)-√(4) d = (2√(3)-2) cm. d = 1.4641016151 units. Area Blue is; Area trapezoid with parallel sides 2 cm and 2√(3) cm respectively and height (2√(3)-2) cm + Area sector with radius 4 cm and angle 30° - Area triangle with height 4 cm and base 4sin30 cm. = (½*(2+2√(3))*(2√(3)-2))+(30π*4*4/360)+(½*4*4sin30) = (½*8)+⅓(4π)-4 = 4+⅓(4π)-4 = ⅓(4π) square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2578138

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a = 40+10 a = 50° a is angle ACB. b = 180-10-50 b = 120° b is angle ADC. (6/sin120) = (c/sin10) c = 1.2030698654 units. c is CD. (6/sin120) = (d/sin50) d = 5.3073115854 units. d is AD. e = 180-120 e = 60° e is angle ADB. f = 180-50-50 f = 80° f is angle ABD. The required length, x (BD) is; (5.3073115854/sin80) = (x/sin40) x = 3.4641016151 units. Again, x is the required length, BD.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2573455

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Let a be the side length of the inscribed square. b² = 2a² b = √(2)a units. b is the radius of the half circle. tanc = a/√(2)a c = atan(1/√(2))° d² = a²+(√(2)a)² d² = a²+2a² d = √(3)a units. e = (2+√(3)a) units. f = 2b f = 2√(2)a f = √(2)a unit. Calculating a. cos(atan(1/√(2))) = (2+√(3)a)/(2√(2)a) 0.8164965809 = (2+√(3)a)/(2√(2)a) 2.3094010767a-√(3)a = 2 0.5773502691a = 2 a = 3.4641016156 units. a = 2√(3) units. Again, a is the side length of the inscribed green square. Area green squ...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2571520

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a = 8+9 a = 17 units. a is the radius of the half circle. b²+8² = 17² b = √(225) b = 15 units. c²+d² = 17² c = √(289-d²) units. e = (17-d) units. f = ½(c) f = ½√(289-d²) unit. g = 17+17-9 g = 34-9 g = 25 units. Observing similar plane shape (right-angled) side length ratios. (17-d) - 25 ½√(289-d²) - 15 Cross Multiply. 15(17-d) = 25*½√(289-d²) 30(17-d) = 25√(289-d²) 6(17-d) = 5√(289-d²) 102-6d = 5√(289-d²) (102-6d)² = 25(289-d²) 10404-1224d+36d² = 7225-25d² 11d²-1224d+3179 = 0 It impl...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2561674

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Notice! The ascribed triangle is equilateral. Calculating the required angle x. a² = 1+3²-2*3cos60 a = √(7) units. (√(7)/sin60) = (1/sinb)  b = 19.1066053509° c = 60-b c = 40.8933946491° d = 180-b-c d = 120° (3/sin120) = (e/sin40.8933946491) e = 2.2677868381 units. f² = 2.2677868381²+3²-2*2.2677868381*3cos40.8933946491 f = 1.9639610121 units. Therefore, the required angle x is; (3/sinx) = (1.9639610121/sin40.8933946491) x = 90°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2541591

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Calculating the area of the blue inscribed rectangle and the area of the ascribed regular pentagon. πr² = 36π r = 6 cm. r is the radius of the inscribed circle. sin(0.5*108) = 6/a a = 7.416407865 cm. b = r+a b = 13.416407865 units. tan36 = 13.416407865/c c = 18.4661012231 cm. d = 2c d = 36.9322024461 cm. d is the width of the inscribed blue rectangle. Therefore; Area Blue Inscribed Rectangle is; 36.9322024461*50 = 1846.6101223052 cm² ≈ 1850 cm² to 3 significan...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2500296

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3² = 2c²-2c²cos108 108° is the single interior angle of the regular pentagon. It implies; 9 = 2.6180339887c² c² = 3.4376941013 c = 1.8541019663 units. c is the side length of the regular pentagon. sin36 = d/c d = 1.8541019663sin36 d = 1.089813792 units. cos36 = e/c e = 1.8541019663cos36 e = 1.5 units. f = 1.5-1 f = 0.5 units. tang = f/d tang = 0.5/1.089813792 g = atan(0.5/1.089813792) g = 24.6453719079° h = 54-g h = 29.3546280921° j = 180-108-h j = 42.6453719...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2497970

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b = (180(12-2))/12 b = 1800/12 b = 150° b is the single interior angle of the regular dodecagon. 2c+2(150) = 360 2c = 60 c = 30° d = 150-15-30 d = 105° 2e+3(150) = 180(5-2) 2e = 540-450 2e = 90 e = 45° f = e-15 f = 30° g = 180-d-f g = 180-105-30 g = 45° It implies; (h/sin105) (2√(2)/sin30) h = 5.4641016151 units. j = 150-15-45 j = 90° k = 180-105-30 k = 45° (l/sin45) = (2√(2)/sin105) l = 2.0705523608 units. l is the side length of the regular dodecagon. m² = 2(2.0705523608)²-2(2.070...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2473303

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sin60 = 18/a a = 20.7846096908 units. tan60 = 18/b b = 10.3923048454 units. c+270+60+30+135 = 180(5-2) c = 540-495 c = 45° d² = 2(18²) d = 18√(2) units. d = 25.4558441227 units. e = a+d e = 20.7846096908+25.4558441227 e = 46.2404538135 units. f = (85+z) units. f is the required length x. Therefore: x = f = (85+z) units. g = (46.2404538135+z) units. tan60 = h/a h = 20.7846096908tan60 h = 36 units. j = (25.4558441227+z) units. j is the required length y. y...
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