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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3061909

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Let AC be 1 unit. Area smaller pentagon is; ½(5)*(1/(2tan(36))) = 1.72047740059 square units. a² = 1²+0.5²-cos108 a = 1.24860602048 units. Where a is CD. (1.24860602048/sin108) = (0.5/sinb) b = 22.38617755917° c = 108-22.38617755917 c = 85.61382244083° Calculating Area E. It is; 0.5*0.5*0.5*1.24860602048sin22.38617755917 = 0.05944103227 square units. (a) Area E ÷ Area Smaller Pentagon to 3 decimal places is; 0.05944103227 ÷ 1.72047740059 = 0.034549...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3058839

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a² = 10²+16²-2*10*16cos120 a = 2√(129) units. Where a is length AC. (3√(129)/sin120) = (16/sinb) b = 37.58908946897° c = 90-b c = 52.41091053103°  d = 60-b d = 22.41091053103° e = 90-d e = 67.58908946897°  f = 90-60 f = 30° cos37.58908946897 = g/10 g = 7.92405815693 units. i = 30+37.58908946897 i = 67.58908946897° tan67.58908946897 = j/7.92405815693 j = 19.2148198619 units. Where j is the radius of the arc. sin67.58908946897 = 19.2148198619/k k = 20....

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3004810

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a = ½(180-150) a = 15° b = 90-15 b = 75° (c/sin45) = (12/sin60) c = 9.79795897113 cm. (12/sin60) = (d/sin45) d = 9.79795897113 cm. e² = 9.79795897113²+12²-24*9.79795897113cos135  e = 20.15631110909 cm. (12/sinf) = (20.15631110909/sin135) f = 24.89609063899° g = 120-24.89609063899 g = 95.10390936101° h = 180-24.89609063899-75 h = 80.10390936101° (9.79795897113/sin80.10390936101) = (i/sin75) i = 9.60704430088 cm. Therefore; Area Shaded Region is; 0.5*9...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2987380

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Let a be the side of the inscribed square. Calculating a. tan60 = a/b b = a/(tan60) cm. Where b is BS = CR. Therefore; 2b+a = 10 2a/(tan60)+a = 10 a((2/tan60)+1) = 10 a = 10(2√(3)-3) cm. a = 4.64101615138 cm. c = ½(10-a) c = 0.5(10-4.64101615138) c = 2.67949192431 cm. Where c = b. d = c+a d = 2.67949192431+4.64101615138 d = 7.32050807569 cm. Where d is BR. e = atan(4.64101615138/7.32050807569) e = 32.37365912705° It implies; Calculating TI, let it be f. tan32.37365912705 = f/g g = f/...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2977546

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Let the inscribed square side be a. tan60 = a/b a = √(3)b cm. Where b is BS = CR It implies; b+√(3)b+b = 10 (2+√(3))b = 10 b = 10(2-√(3)) cm. Therefore; a = √(3)*10(2-√(3)) a = 10(2√(3)-3) cm. a = 4.64101615138 cm. sin60 = 10(2√(3)-3)/c c = 20(2-√(3)) cm. Where c is CQ. d² = 10²+(10(2-√(3)))²-20*10(2-√(3))cos60 d = 8.96575472168 cm. Where d is AS. (8.96575472168/sin60) = (10(2-√(3))/sine) e = 15° Where e is angle PAS. f = 180-60-15 f = 105° Where f is angle BSA. g = 105-90 g = 15° Whe...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2974204

mathematics-question-and-solution_1739323824vaVWsU.jpeg
Let the inscribed square side be a. tan60 = a/b a = √(3)b cm. Where b is BS = CR It implies; b+√(3)b+b = 10 (2+√(3))b = 10 b = 10(2-√(3)) cm. Therefore; a = √(3)*10(2-√(3)) a = 10(2√(3)-3) cm. a = 4.64101615138 cm. sin60 = 10(2√(3)-3)/c c = 20(2-√(3)) cm. Where c is CQ. d² = 10²+(10(2-√(3)))²-20*10(2-√(3))cos60 d = 8.96575472168 cm. Where d is AS. (8.96575472168/sin60) = (10(2-√(3))/sine) e = 15° Where e is angle PAS. f = 180-60-15 f = 105° Where f is angle BSA. g = 105-90 g = 15° Whe...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2940636

mathematics-question-and-solution_1737510024UFRouO.jpeg
Let AC be 1 unit. Area smaller pentagon is; ½(5)*(1/(2tan(36))) = 1.72047740059 square units. a² = 1²+0.5²-cos108 a = 1.24860602048 units. Where a is CD. (1.24860602048/sin108) = (0.5/sinb) b = 22.38617755917° c = 108-22.38617755917 c = 85.61382244083° Calculating Area E. It is; 0.5*0.5*0.5*1.24860602048sin22.38617755917 = 0.05944103227 square units. (a) Area E ÷ Area Smaller Pentagon to 3 decimal places is; 0.05944103227 ÷ 1.72047740059 = 0.034549...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2940069

mathematics-question-and-solution_1737472714NkUT6I.jpeg
a² = 10²+16²-2*10*16cos120 a = 2√(129) units. Where a is length AC. (3√(129)/sin120) = (16/sinb) b = 37.58908946897° c = 90-b c = 52.41091053103°  d = 60-b d = 22.41091053103° e = 90-d e = 67.58908946897°  f = 90-60 f = 30° cos37.58908946897 = g/10 g = 7.92405815693 units. i = 30+37.58908946897 i = 67.58908946897° tan67.58908946897 = j/7.92405815693 j = 19.2148198619 units. Where j is the radius of the arc. sin67.58908946897 = 19.2148198619/k k = 20....

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2899870

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#Sir Mike Ambrose is the author of the question. Let the side length of the regular Pentagon be 1 unit. Therefore Area Pentagon is; (5*1)÷(4tan(180/5)) = 5/(4tan36) space units. = 1.72047740059 space units. Area star is; 5(area triangle with height 0.61803398875 units and base 0.61803398875sin36 units) + Area regular Pentagon with side 0.38196601125 units. = 5(0.5*0.61803398875*0.61803398875sin36) + ((5*0.38196601125²)/(4tan(180/5))) = 0.56128497072 + 0.25101426986 = 0.812...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2886119

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Sir Mike Ambrose is the author of the question. Side length of the square is; Let it be x. x = √((16*34)-144) x = √(400) x = 20 cm. Therefore; Area Orange exactly in decimal cm² is; Half area square with side 20 cm - Area triangle with height 7.88405797101 cm and base ⅓(32sin(2atan(3/5))) cm - Area triangle with height 8.74642784227 cm and base (12sin(atan(5/3))) cm = ½(20²) - ½(7.88405797101*⅓(32sin(2atan(3/5)))) - ½(8.74642784227*(12sin(atan(5/3)))) = 200 - 37.10144927...
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