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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1933091

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Notice; The diameter of the semicircle is 8 units. Therefore radius is 4 units. tan30 = a/2 a = ⅔(√(3)) units. b = 3a b = 2√(3) units. c² = 2²+(⅔(√(3)))² c = (4/√(3)) units. d = (2/√(3))+c d = 2√(3) units. e² = 4²+4²-2*4*4cos120 e = 4√(3) units. f = atan((4/√(3))/2) f = atan(2/√(3)° g² = 4²+4²-2*4*4cos(180-2atan(2/√(3))) g = (8√(21)/7) units. h = (atan(2/√(3))-30)° Area Maroon exactly in square units is; Area triangle with height (8√(21)/7) units and base 4√(3)sin(atan(2/√(3))-30)...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1932925

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Let alpha be x. 6x = 60 x = 10° Alpha is 10° Check accuracy. x+x+2x+8x+6x = 18x And x = 10 Therefore; 18x = 18*10 = 180° which is the sum of the interior angles of the ascribed triangle.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1931994

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a² = 2*12² a = 12√(2) cm. Where a is AC = AE. b = 180-45 b = 135° Where b is angle EAB. c² = (12√(2))²+12²-24*12√(2)cos135 c = 12√(5) cm. Where c is BE. (12√(5)/sin135) = (12√(2)/sind) d = 26.56505117708° Where d is angle ABE. e = atan(½)° Where e is angle BCF. f = 90+26.56505117708  f = 116.56505117708° Where f is angle CBG. g = 180-116.56505117708-atan(½) g = 36.86989764584° Where g is angle BGF. (12/sin36.86989764584) = (h/sin116.56505117708) h = 17.88854382 cm. Where h is CG. i² =...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1931302

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a = 90+60 a = 150° Where a is angle ADF b² = 12²+12²-2*12*12cos150 b = 23.18221983094 cm. Where b is AF. c = ½(180-150) c = 15° Where c is angle DAF = angle DFA. d = 90-15 d = 75° Where d is angle FAB. Notice; Angle ABG is 45° e = 180-45-75 e = 60° Where e is angle AGB. Calculating Length BG. Let it be f. (12/sin60) = (f/sin75) f = (12sin75)/sin60 f = (2√(6)+6√(2)) cm. f = 13.3842608598 cm. Therefore; BG = 13.3842608598 cm.

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/1929550

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The radius of the circle is 6 cm. a = atan(2)° Where a is angle AOB = angle COE. Angle BOE = 180-atan(2) = 116.56505117708° Angle OBE = ½(180-116.56505117708) = 31.71747441146° b = 90+31.71747441146 b = 121.71747441146° Where b is angle ABE. (12/sin31.71747441146) = (c/sin121.71747441146) c = 19.416407865 cm. Where c is AE. sin(atan2) = d/19.416407865 d = 17.366563146 cm.  cos(atan2) = e/19.416407865 e = 8.683281573 cm. f = 12-8.683281573   f = 3.316718427 c...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1928279

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Let r = 1 units. Let the ascribed half circle's radius be x. Therefore; a = (x-2) units. b = ½(x-2) units. c = x-b c = x-½(x-2) c = ½(x+2) units. d = 2a d = 2(x-2) d = (2x-4) units. Therefore x² = c²+d² x² = (½(x+2))²+(2x-4)² x² = ¼(x²+4x+4)+4x²-16x+16 4x² = x²+4x+4+16x²-64x+64 13x²-60x+68 = 0 Therefore, solving the quadratic equation. x ≠ 2 units. x = (34/13) units. x = 2.61538461538 units. Where x is the radius of the ascribed half circle (R). Notice, r , radius of the inscribed sma...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1928097

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Let the side length of square ABCD be a. Area Square ABCD will be a² Calculating a² Notice, he three inscribed plane shapes, two congruent right-angled triangle and a parallelogram has equal area. Therefore; ½(3ab) = a² b = (2a/3) --- (1).  Where b is the base length of the two congruent inscribed right-angled triangle. ab + √(a²+b²) = a² --- (2). Substituting (1) in (2). a(2a/3) + √(a²+(2a/3)²) = a² (2a²/3)+√(a²+(4a²/9)) = a² (13a²/9) = (a²/3)² (13a²/9) = (a⁴/9) 13 = a² a² is the are...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1927886

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Let the regular pentagon side be 1 unit. a = ⅕(180*3) a = 108° a is an angle of the regular pentagon. b = 108-36 b = 72° c = 180-45-72 c = 180-117 c = 63° (d/sin63) = (1/sin45) d = 1.26007351067 units. cos72 = e/1 e = 0.30901699437 units. f = 2e+1 f = 2(0.30901699437)+1 f = 1.61803398875 units. g = f-d g = 1.61803398875-1.26007351067 g = 0.35796047808 units. h² = 0.35796047808²+0.5²-2*0.35796047808*0.5cos36 h = 0.2975560347 units. Calculating x. (0.5/sinj) = (0.2975560347/sin36) sinj...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1926872

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Let the inscribed regular pentagon side be 1 unit. Area small/inscribed pentagon is; 0.5*5(1/(2tan(180/5))) = 1.72047740059 square units. Calculating Shaded Area. a² = 2-2cos(60+108) a = 1.98904379074 units. b = ½(180-108-60) b = 6° c = 60-2(6) c = 48°   sin60 = d/1 d = 0.86602540378 unit. tan24 = e/0.86602540378 e = 0.38557935197 unit. f = 2e f = 0.77115870394 units. g = 0.5(1-f) g = 0.11442064803 unit. cos24 = 0.86602540378/h h = 0.94798282509 unit. Area Shaded is; 48*π*1.98904379...
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