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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1919162

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Let AB be 1 unit. (1/sin40) = (BC/sin70) BC = 1.46190220008 units. (1.46190220008/sin120) = (AE/sin40) AE = 1.08506357513 units. (1.08506357513/sin40) = (CE/sin20) CE = 0.57735026919 units. BE = 1.46190220008-0.57735026919 BE = 0.88455193089 units. (BD)² = 0.88455193089²+0.57735026919²-2*0.57735026919*0.88455193089cos60 BD = 0.77786191343 units. Therefore angle alpha (angle BDE) is; Let it be x. (0.88455193089/sinx) = (0.77786191343/sin60) x = 80° Angle alpha (required angle) = 80°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1918922

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a² = 12²-6² a = 6√(3) cm. Where a is BD. b = ½(a) b = 3√(3) cm. Where b is BG = DG. c = atan(6/(3√(3)) c = 49.10660535087° Where c is angle AGD = angle BGE. d = 60-49.10660535087 d = 10.89339464913° Where d is angle BAE. e = 180-60-10.89339464913 e = 109.10660535087° Where e is angle BEA. (12/sin109.10660535087) = (f/sin60) f = 10.99818166789 cm. Where f is AE. (12/sin109.10660535087) = (g/sin10.89339464913) g = 2.4 cm. Where g is BE. h = 12-2.4 h = 9.6 cm. Where h is CE. i² = (6√(3))²...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1916952

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Let AB be 5 units. Radius, r of the inscribed circle is; 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 units. g = 52.5+0.5(45) g = 75° sin(...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1916500

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AD = √(40²-20²) AD = 20√(3) units. AE = DE = ½(AD) = 10√(3) units. Angle BED is; = atan(20/(10√(3))) = atan(⅔√(3))° = 49.10660535087° Angle DAC is; = 30° (EF)² = 300+625-20√(3)*25cos30 EF = 5√(7) units. (5√(7)/sin30) = (25/sina) a = 70.89339464913° a is angle DEF. Therefore required angle is; Angle BED + Angle DEF = 49.10660535087+70.89339464913 = 120°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1916431

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Let AD = 2AB. For AB = 2 units, AD = BC = 4 units. Angle APD is; = atan(4)° PD = √(4²+1)  PD = √(17) units. Considering similar triangles APD and QCD ratios. 1 = a √(17) = 2 Crossing Multiply. a = (2/√(17)) units. a = 0.48507125007 units. a is DQ. PQ = √(17)-0.48507125007  PQ = 3.63803437554 units. Angle BPQ = 180-atan(4) = 104.03624346793° Calculating QB. Notice, BC is 4 units. (QB)² = 3.63803437554²+1-2*3.63803437554cos104.03624346793 QB = 4 unit. It implies;  BC = QB Proved....

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1916258

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Calculating R. (5+R)² = (10-R)²+(5-R)² 25+10R+R² = 100-20R+R²+25-10R+R² 20R = 100-20R+R² R²-40R+100 = 0 Therefore; R ≠ 20+10√(3) units. R = 20-10√(3) units. R = 2.67949192431 units. Calculating r. (5-r)² = r²+a² 25-10r+r² = r²+a² a = √(25-10r) units. Therefore; (5+r)² = (5-r)²+ (5+√(25-10r))² 25+10r+r² = 25-10r+r²+25+10√(25-10r)+25-10r 20r = 25+10√(25-10r)+25-10r 30r-50 = 10√(25-10r) (3r-5)² = 25-10r 9r²-30r+25 = 25-10r 9r² = 20r 9r = 20 r = ⅑(20) units. r = 2.22 units. It implies; r ÷...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1916187

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b² = 12²+12² b² = 144+144 b² =288 b = 12√(2) cm. Where b is AE = AC. c² = 12²+6² c = 6√(5) cm. Where c is AM. d = b-c d = (12√(2)-6√(5)) d = 3.55415488348 cm. Where d is EM. e = 45-atan(1/2) e = 18.43494882292° Where e is angle MAC. f = 45+18.43494882292  f = 63.43494882292° Where f is angle EAB. sin63.43494882292 = g/(12√(2)) g = 15.17893276881 cm. cos63.43494882292 = h/(12√(2)) h = 7.5894663844 cm. i = 12-7.5894663844  i = 4.4105336156 cm. j = atan(15.17893276881/4.4105336156) j = 73...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1916055

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Let the AC = AB = a (diameter of the semi circle). Therefore; sin22.5 = AD/AC AD = asin22.5 units. cos22.5 =CD/AC CD = acos22.5 units. Calculating a. 84+25+½(acos22.5)(asin22.5) = ½(a²) 109+⅛√(2)a² = ½(a²) 109 = ½(a²)-⅛√(2)a² 109 = ⅛(4-√(2))a² 872 = (4-√(2))a² a = √(872/(4-√(2))) a = 18.36377300632 units. It implies; Green Area is; 0.5a²-84-25 = 0.5*18.36377300632²-84-25 = 59.61407951382 square units. Or = 0.5*18.36377300632²*sin22.5*cos22.5 = 59.6140...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1915865

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Let a be the radius of the inscribed circle. Calculating a. tan30 = a/6 a = 2√(3) cm. b = inscribed circle diameter. b = 4√(3) cm. 12² = c²+6² c = 144-36 c = √(108) c = 6√(3) cm. d = 6√(3)-4√(3) d = 2√(3) cm. Where d is AD. e² = 12²+(2√(3))²-2*12*2√(3)cos30 e = 2√(21) cm. Where e is BD. (2√(21)/sin30) = (12/sinf) f = 139.10660535087° Where f is angle BDA. g = 180-30-139.10660535087  g = 10.89339464913° Where g is angle ABD. h = 60-10.89339464913  h = 49.10660535087° Where h is angle FB...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1915750

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The single interior angle of the ascribed regular hexagon is; ⅙(180*4) = 120° a = 360-90-120-(120-53) a = 150-120+53 a = 83° b = 360-120-53-53 b = 240-106 b = 134° It implies; The required angle ? is; 360-46-134-83 = 360-180-83 = 180-83 = 97°
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