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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2876016

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Let the side of the regular pentagon be 1 unit. Notice; The regular pentagon side is equal the regular heptagon side. a = ⅐(180*5) a = ⅐(900)°, the single interior angle of the regular heptagon. b = 2a b = ⅐(450)° tan(⅐(450)) = c/0.5 c = 1.03826069829 units. cos(⅐(450)) = 0.5/d d = 1.15238243548 units. Where d is the radius of the ascribed circle. e = 180-⅐(900) e = ⅐(360)° tan18 = 0.5/f f = 1.53884176859 units. g² = 1.15238243548²-0.5² g = 1.03826069828 u...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2874400

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Let AB be 5 units. Radius, r of the inscribed circle is; r  = 5*(2/5) r = 2 units. a = ⅛(180*6) a = 135° 2b² = 25 b = ½(5√(2)) units. c = 180-0.5(135)-45 c = 67.5° sin22.5 = ½(5√(2))/d d = 9.23879532511 units. Where d is the side length of the regular triangle (purple area). Area purple is; 0.5*9.23879532511*9.23879532511sin60 = 36.95994598698 square units. Calculating Area Shaded. e = 360-135-45-67.5-60 e = 52.5° f² = 50-50cos135 f = 9.23879532511...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2874392

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Sir Mike Ambrose is the author of the question. Let the side length of the square be 2 units. Therefore area ABD (equilateral triangle) is; Area triangle with height 2 unit and base (2sin60) units. = ½*2*2sin60 = √(3) square units. Area purple is; Area triangle with height 2√(5-2√(3)) units and base (√(3)/3sin(atan(⅓(2+√(3)))))sin(atan(⅓(2+√(3)))+atan(½(√(3)+1))) units. = ½(2√(5-2√(3)))(√(3)/3sin(atan(⅓(2+√(3)))))sin(atan(⅓(2+√(3)))+atan(½(√(3)+1))) = ½*(2√(5-2√(3))...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2872619

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18² = 14²+16²-2*14*16cosa 448cosa = 14²+16²-18² a = acos((14²+16²-18²)/448) a = 73.39845040098° (18/sin73.39845040098) = (14/sinb) b = 48.18968510422° c = 180-48.18968510422-73.39845040098 c = 58.4118644948° cotd = cota+cotb+cotc cotd = 1.80748828181  d = acot(1.80748828181) d = 28.95373454979° e = 58.4118644948-d e = 29.45812994501° f = 180-28.95373454979-73.39845040098 f = 77.64781504923° (14/sin77.64781504923) = (g/sin73.39845040098)  g = 13.73434221224 units....

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2872525

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Let the inscribed purple regular heptagon side be 1 unit. a = ⅐(180*5) a = ⅐(900)° b = 180-⅐(900)) b = ⅐(360)° c = 180-2(⅐(360)) c = ⅐(540)° (1/sin⅐(540)) = (d/sin⅐(360)) d = 0.8019377358 unit. e = 2(0.8019377358)+1 e = 2.60387547161 units. f = 90-½(⅐(540)) f = ⅐(360)° cos⅐(360) = g/2.60387547161 g = 1.62348980186 units. h = 2g h = 3.24697960372 units. Where h is the base of the ascribed quadrilateral. cos(180/7) = i/0.8019377358 i = 0.72252093395 unit....

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2870872

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Let the side length of the regular pentagon be 1 unit. tan36 = a/0.5 a = 0.363271264 unit. b² = 0.363271264²+0.5² b = 0.61803398875 unit. tan72 = c/0.5 c = 1.53884176859 units. Area Green is; 0.5sin108+0.5*1.53884176859*0.5-0.5*0.5*0.363271264-0.5*0.61803398875*0.61803398875sin36 = 0.65716389015 square units. Calculating Area Blue. Let d be the radius of the inscribed blue circle. Calculating d. e = 0.5(90+36) e =63° (d/tan63) + (d/tan45) = a d = 0.36327126...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2870862

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Sir Mike Ambrose is the author of the question. Let the single side length of the square be 2 units. Therefore; Area Square is; 2² = 4 square units. Area Green is; Area triangle with height √(3) units and base (4√(3)-6) units. = ½(√(3)*(4√(3)-6)) = √(3)(2√(3)-3) = (6-3√(3)) square units. It implies; Area Green ÷ Area Square exactly in its simplest form is; (6-3√(3)) ÷ 4 = ¼(6-3√(3)) It implies; Angle Alpha, Q, exactly in its simplest form is;...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2869662

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Let the side length of the regular heptagon be 1 unit. a = ⅐(180*5) a = ⅐(900)° b = 0.5(360-2(900/7)) b = ⅐(360)° c = 2cos(360/7)+1 c = 2.24697960372 units. d² = 2-2cos(90/7) d = 1.8019377358 units. Where d is the radius of the arc. Area Orange is; 0.5*2.24697960372*1sin(360/7)+0.5*sin(900/7) = 1.26929543856 square units. Let e be alpha. Calculating e. e = 0.5((900/7)-(360/7)) e = ⅐(270)° (1.8019377358/sin(270/7)) = (2.24697960372/sinf) f = 128.96966294...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2869571

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Sir Mike Ambrose is the author of the question. Area S, exactly in square units is; Area triangle with height 2 units and base (2sin120) units + Area triangle with height 2 units and base 2√(2) units - Area sector with radius 2 units and angle (30+atan(√(2)))° = (½*2*2sin120) + (½*2*2√(2)) - (4π*(30+atan(√(2)))÷360) = √(3) + 2√(2) - (π(30+atan(√(2)))/90) square units. = 4.56047793232 - 2.95783078745 = 1.60264714487 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2868366

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a² = 11*11+10*10-2*11*10cos50 a = 8.92113926968 cm. (8.92113926968/sin50) = (10/sinb) b = 59.16920318624° c = 180-50-59.16920318624 c = 70.83079681376°  d = 0.5a d = 4.46056963484 cm. e² = 4.46056963484²+5.5²-2*4.46056963484*5.5cos59.16920318624 e = 5 cm. (5/sin59.16920318624) = (5.5/sinf) f = 70.83079681376° g² = 4.46056963484²+5²-2*4.46056963484*5cos70.83079681376 g = 5.5 cm. (5.5/sin70.83079681376) = (5/sinh) h = 59.16920318624° i = 180-59.16920318624-70.83...
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