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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1984724

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(3/sin108) = (EG/sin18) EG = 0.9747590887 units. (3/sin108) = (EF/sin54) EF = 2.55195242506 units. AF = 4-2.55195242506 AF = 1.44804757494 units. sin18 = a/1.44804757494  a = 0.44747130932 units.  cos18 = b/1.44804757494 b = 1.37717508205 units. b is the height of the trapezoid. Let BJ be x. c = (2-x) units. d² = 1.44804757494²+2²-4*1.44804757494cos108 d = 2.80833171413 units. (2.80833171413/sin108) = (2/sine) e = 42.63384501172° f = 72-42.63384501172   f = 29.36615498828° g² = 3²+2...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1982892

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Let the radius of the large circle be 1 unit. Diameter of the large circle = 2 units. sin60 = 2/a a = ⅓(4√(3)) units. Where a is the side length of triangle ABC. b² = 2-2cos b = √(3) units. c = a-b c = ⅓(√(3)) units. (d/sin50) = (4√(3)/(3sin100)) d = 1.79639514045 units. e² = (⅓(2√(3)))²+1² e = ⅓√(21) units. 1² = (⅓√(21))²+f²-2*⅓√(21)f cos(atan(3/(2√(3))-10) f² - 2.62161244054f + 1.33333333333 = 0  f = 1.93113 units. g² = 1.93119²+(⅓(4√(3)))²-2*1.93119(⅓(4√(3)))cos10 g = 0.5277832777...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1958336

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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 = 20.78460969082 units. l² = (2√(129)...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1954045

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Let the square side be 2 units. Notice; a+b = c --- (1). 2√(2) = c+a 2√(2)-a = c --- (2). a² = 2d² d = (a/√(2)) units. e = 2-d e = (2-(a/√(2))) units. b² = (2-(a/√(2)))²+(a/√(2))² b² = 4-(4a/√(2))+½(a²)+½(a²) b = √(4-(4a/√(2))+a²) --- (3). At (1). a+b = c Substituting (3) in (1). a+√(4-(4a/√(2))+a²) = c --- (4). Substituting (2) in (4). a+√(4-(4a/√(2))+a²) = 2√(2)-a √(4-(4a/√(2))+a²) = 2√(2)-2a 4-(4a/√(2))+a² = 8-8√(2)a+4a² 3a²-8√(2)a+2√(2)a+4 = 0 3a²-6√(2)a+4 = 0 3a²-8.48528137424a+4...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1942103

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a² = 6²+8²-2*6*8cos30 a = 4.10628314132 cm. a = DE. (4.10628314132/sin30) = (8/sinb) b = 103.06431342951° b = Angle ADE. Angle AED = 180-30-103.06431342951  = 46.93568657049° tan30 = c/6 c = 3.46410161514 cm. c = CH. d = 8-2*3.46410161514  d = 1.07179676972 cm. e = 1.07179676972+3.46410161514 e = 4.53589838486 cm. f = atan(6/4.53589838486)  f = 52.91133691903° g = 90-52.91133691903  g = 37.08866308097° h = 37.08866308097-30 h = 7.08866308097° i = 60+46.93568657049 i = 106.93568657049°...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1935400

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i = k  And i = 180-101-39 = 40° Therefore; k = 40° While j = 101° Or Calculating i. i = 90-11-39 i = 90+50 i = 40° a = 180-2(11) a = 180-22 a = 158° b = 360-a b = 360-158 b = 202° Calculating j. j = ½(b) j = ½(202) j = 101° Calculating k. k+j+39 = 180 k+101+39 = 180 k = 180-140 k = 40°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1935205

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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*9.60704430088*9.79795897113sin95....

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1930117

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a² = 3²+4² a = 5 cm. Where a is AC. b = atan(4/3)° Where b is angle DAC. c = 60+atan(4/3) c = 113.13010235416° Where c is angle EAC. d² = 3²+5²-2*3*5cos113.13010235416 d = 6.76643256752 cm. Where d is CE. (6.76643256752/sin113.13010235416) = (3/sine) e = 24.06202137986° Where e is angle ACF. f = atan(3/4)-24.06202137986  f = 12.80787626598 cm. Where f is angle DCE. tan(12.80787626598) = g/4 g = 0.90935597101 cm. Where g is DF. cos12.80787626598 = 4/h h = 4.10206390516 cm. Where h is CF....

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1881048

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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/1877260

mathematics-question-and-solution_1686339360ZP9UwP.jpeg
AD = √(12²+6²) AD = 6√(5) cm. Angle AMD = atan(2√(5))° Angle EBM = (90-ata(2√(5)))° sin(90-ata(2√(5))) = EM/9 EM = 1.96396101212 cm. Angle CME = 180-atan(2√(5)) = 102.60438264838° (CE)² = 9+1.96396101212²-6*1.96396101212cos102.60438264838 CE = 3.92792202425 cm. (3.92792202425/sin102.60438264838) = (1.96396101212/sina) a = 29.20593224736° Where a is angle ECM. cos29.20593224736 = 6/CF CF = 6.87386354243 cm. EF = CF-CE  EF = 6.87386354243-3.92792202425 EF = 2.94594151818 cm. Therefore;...
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