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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2067532

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20² = 12²+a² a² = 400-144 a = √(256) a = 16 cm. b = 16-9 b = 7 cm. c = atan(3/4)° Let Alpha be d. d = ½(c) d = 0.5atan(¾)° tan(0.5atan(¾)) = e/16 e = ⅓(16) cm. tan(0.5atan(¾)) = f/7 f = ⅓(7) cm. g = 12-⅓(16) g = ⅓(36-16) g = ⅓(20) cm. cos(0.5atan(¾)) = 7/h h = 7.37864787373 cm. cos(0.5atan(¾)) = 16/j j = 16.86548085423 cm. k = 16.86548085423-7.37864787373 k = 9.4868329805 cm. l = ½(k) l = 4.74341649025 units. ¾ = m/7 m = ¼(21) cm. n = ¼(21)-⅓(...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2066258

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Sir Mike Ambrose is the author of the question. a² = 32-32cos120 a = 4√(3) units. b = 2a b = 8√(3) units. c² =32-32cos150 c = 7.72740661031 units. sin30 = d/4 d = 2 units. sin60 = e/4 e = 2√(3) units. tan15 = 2√(3)/f f = (6+4√(3)) units. g =f-(8√(3)-2) g = (8-4√(3)) units. It implies; Area yellow rectangle is; 4*(8-4√(3)) = (32-16√(3)) square units. Area yellow pentagon is; 2(area triangle with height 4 units and base 4sin120 units) + Area triangle with...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2064854

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Calculating x. ½(x+2)(x-2) = 30 x²-4 = 60 x² = 64 x = 8 m. Therefore; AB = 8-2 = 6 m. BC = 8+2 = 10 m. It implies; L² = 6²+10² L² = 36+100 L² = 136 L = √(136) L = √(2*2*34) L = 2√(34) m.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2064458

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Sir Mike Ambrose is the author of the question. a = atan(¼)° b² = 12²+3² b = 3√(17) cm. c = ½(atan (4/3))° Let the radius of the inscribed circle be r. Calculating r. tan(½(atan(4/3))) = (r/(9-r)) ½ = (r/(9-r)) 2r = 9-r r = 3 cm. d = (atan(¼) + atan(¾))° tan(atan(¼)) = e/9 e = ¼(9) cm. f² = (¼(9))²+9² f = ¼(9√(17)) cm. g² = (9-(9/4))²+6² g = ¼(3√(145)) cm. h² = (2√(5))²+(3√(5))² h = √(65) cm. (b) Angle theta exactly in decimal is; 180-atan(8/9)-atan(2)-atan(⅔) = 41.24144431453° (a)...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2064448

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Let the side length of the regular pentagon be a. b² = a²+(0.5a)²-2*a*(0.5a)cos108 b = 1.24860602048a units.  Calculating angle AGE. Let it be c. (1.24860602048a/sin108) = (a/sinc) c = 49.6138224408° d = 180-2*49.6138224408  d = 80.7723551184° d is the angle of the circular arc EFG with radius 5 units. e² = 50-50cos80.7723551184 e = 6.47936163294 units. Calculating a, side length of the regular pentagon. e = b 6.47936163294 = 1.24860602048a a = 6.47936163294/1.24860602048 a = 5.1892763...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2063981

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a² = 5-4cos120 a = √(7) units. (√(7)/sin120) = (2/sinb) b = 40.89339464913° c = 180-2*40.89339464913 c = 98.21321070174°  Calculating r. 0.5*r(r-1)sin98.21321070174 +(3/2) = (98.21321070174π*r²)/360 0.49487165931r²-0.49487165931r+(3/2) = 0.85707194785r²  0.36220028854 r²+0.49487165931r-(3/2) = 0 0.36220028854 r²+0.49487165931r-1.5 = 0 Solving the quadratic equation to get r. r = 1.46349 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2063142

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Let the side length of the regular pentagon be 1 unit. c² = 2-2cos108 c = 1.61803398875 units. (d/sin36) = (1/sin108) d = 0.61803398875 units. e = c-d e = 1 unit. Calculating a. sin72 = a/1 a = 0.9510565163 units. Calculating b. sin72 = b/0.61803398875 b = 0.58778525229 units. It implies; (a/b)-(b/a) = (0.9510565163/0.58778525229)-(0.58778525229/0.9510565163) = 1

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2062073

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Sir Mike Ambrose is the author of the question. Let the two equal length be 2 units each. 2 = 2/a a = 1 unit. sin(atan(½)) = b/3 b = ⅕(3√(5)) units. c = 2b c = ⅕(6√(5)) units. Where c is the square side. d² = 2²+1² d = √(5) units. e = ⅕(6√(5))-d e = ⅕(√(5)) unit. sin(atan2) = f/1 f = ⅕(2√(5)) units. cos(atan2) = g/1 g = ⅕(√(5)) unit. h = ⅕(3√(5))-⅕(√(5)) h = ⅕(2√(5)) units. 3 = (⅕(2√(5)))/i i = (2√(5))/15 unit. Therefore; Length FG exactly is; d-g-i = √(5)-⅕(√(5))-(2√(5))/15 = ⅓(2√(...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2060739

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Calculating equation of the curve. 2a² = 4² a = 2√(2) b = (4+2√(2)) b is the maximum turning point. Therefore considering points; (0, 0) and ((4+2√(2)), (4+2√(2))) y-(4+2√(2)) = c(x-(4+2√(2)))² --- (1). Substituting (0, 0) in (1). -(4+2√(2)) = c(-(4+2√(2)))² -(4+2√(2)) = (24+16√(2))c -(2+√(2)) = (12+8√(2))c c = -¼(2-√(2)) It implies; Substituting c in (1). y = -¼(2-√(2))(x-(4+2√(2)))²+(4+2√(2)) Therefore; y = 2x-¼(2-√(2))x² Area Shaded Exactly in square units is; Area trapezoid with...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2059473

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Sir Mike Ambrose is the author of the question. Aqua total in cm² is; tan30 = a/10 a = ⅓(10√(3)) cm. sin60 = 2/b b = ⅓(4√(3)) cm. c = a+b c = ⅓(14√(3)) cm. tan30 = d/2 d = ⅓(2√(3)) cm. e = c-d e = 4√(3) cm. Where e is the side length of the green inscribed regular hexagon. f = (120-atan(7√(3)/15))° g² = c²+10² g = ⅓(4√(93)) cm. h = ⅔(√(3)) cm. i = 6-htan60 i = 2 cm. j = 6-i j = 4 cm. k² = h²+2² k = ⅓(4√(3)) cm. It implies; Total Area Aqua exactly in cm² is; Area triangle with he...
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