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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1854583

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BC = √(3²+4²) BC = 5 cm. tana = (4/3) a = atan(4/3)° tan(½(atan(4/3))) = AF/3 AF = 3tan(½(atan(4/3))) AF = 1.5 cm. b = 90-½(atan(4/3)) b = 63.43494882292° Where b is angle AFE. sin63.43494882292 = AE/1.5 AE = 1.3416407865 cm. (BE)² = 3²-1.3416407865² BE = 2.683281573 cm. Calculating Length CE. (CE)² = 5²+2.683281573²-2*5*2.683281573cos(0.5atan(4/3))  CE = 2.86356421266 cm. (2.683281573/sinc) = (2.86356421266/sin(0.5atan(4/3))) c = 24.77514056879° Where c is angle BCE. Angle ECF = atan...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1854240

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Calculating r, radius of the circle. r² = (r-3)²+4² r² = r²-6r+9+16 6r = 25 r = (25/6) units. a = r-3 a = (25/6)-3 a = (25-18)/6 a = (7/6) units. tanb = (7/6)/4 b = atan(7/24)° c = 180-b c = (180-atan(7/24)° c = 163.73979529169° Therefore; AB Length is; (AB)² = (25/6)²+3²-2(25/6)*3cos163.73979529169 (AB)² = 1813/36 AB = √(1813/36) AB = 7√(37)/6 units. AB = 7.09655628535 units. Or  Length AB is; (AB)² = (r-3)²+7² And r = (25/6) units. (AB)² = ((25/6)-3)²+49 (AB)² = (7/6)²+49 (AB)² = (49...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1852092

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Let r be the radius of the semi circle. Calculating r. √(r²-49)+√(r²-36) = 12 r²-49 = (12-√(r²-36))² r²-49 = 144-24√(r²-36)+r²-36 24√(r²-36) = 144+13 24√(r²-36) = 157 r²-36 = (157/24)² r² = (157/24)²+36 r² = 45385/576 r = √(45385/576) r = √(45385)/24 units. r = 8.87656480728 units  Green Area is; Area semi circle with radius 8.87656480728 units - Area inscribed rectangle with length and width 12 units and 6 units respectively. = 0.5(45385/576)π-(12*6) = (45385/1152)π-72 = (45385π-82944)...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1851945

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Angle DAG = 60-45 = 15° Angle DAG = Angle DCF  = 15° Angle CDF = 90-15 = 75° Angle ADG = 90-75 = 15° It implies; Triangle ADG is an isosceles. Calculating DG. cos15 = 6/(DG) DG = 6/cos15 DG = 6(√(6)-√(2)) cm. DG = 6.21165708246 cm.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1851891

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tanx = 2/a a = 2/tanx ----- (1). b = a+2 b = (2/tanx) + 2 ----- (2). Notice; a = the short adjacent length of the right-angled triangle. b = the long adjacent length of the right-angled triangle. Let tanx = c. Calculating the hypotenuse of the right-angled triangle. (2/c)-2+(2/c) = (4/c)-2 = (4-2c)/c units. It implies; Using sum of areas derived that are inscribed the ascribed triangle to equal the area of the ascribed triangle to find c. 2(2/c)+2(2c+2)/c+2(4-...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1851676

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Let the height of the triangle be a. Let angle alpha be b. tanb = 3/a ----- (1) tan2b = 8/a ----- (2) And tan2b = 2tanb/(1-tan²b) Therefore; 2tanb/(1-tan²b) = 8/a ----- (3) Substituting (1) in (3). (2*(3/a))/(1-(3/a)²) = 8/a (6/a)/(1-(9/a²)) = 8/a (6/a)/((a²-9)/a²) = 8/a (6a/(a²-9)) = 8/a 6a² = 8a²-72 2a² = 72 a² = 36 a = 6 units. Notice, a is the height of the triangle. Therefore x is; √(6²+8²)  = √(100) = 10 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1850096

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Please, move a step right to view the analyzed figure. Notice; ABC triangle is Isosceles. a² = 3²-2² a = √(5) units. Where a is AD. tanb = ½(√(5))° b = atan(½(√(5)))° b = 48.18968510422° Where b is angle ACB. c = 0.5(180-48.18968510422) c = 65.90515744789° Where c is angle ABC. d = 90-65.90515744789  d = 24.09484255211° Where d is angle BAD cos65.90515744789 = e/3 e = 1.22474487139 units. Where e = x = BF. tan24.09484255211 = f/1.22474487139 f = 0.54772255751 u...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1849189

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Let the ascribed quarter circle be 2 units. 2a² = 2² a² = 2 a = √(2) units. b = (2-√(2)) units. c² = (√(2))²+(2-√(2))² c² = 2+4-4√(2)+2 c² = 8-4√(2) c = √(8-4√(2)) units. d² = √(8-4√(2))²-√(2)² d² = 8-4√(2)-2 d² = 6-4√(2) d = √(6-4√(2)) units. tane = √(2)/(√(6-4√(2))) e = atan(√(2)/(√(6-4√(2)))) e = 67.5° f = 180-45-67.5 f = 67.5° tan67.5 = 2/g g = 0.82842712475 units. h = 2-0.82842712475  h = 1.17157287525 units. Therefore; Area Triangle S is; 0.5*2*0.82842712475 = 0.82842712475 sq...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1849033

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Let the square side be 2 units. a = √(2²+1²) a = √(5) units. b = atan(2)° c = atan(½)° d² = 8-8cos150 d = 3.86370330516 units. e = 0.5d e = 1.93185165258 units. f = 90-15-atan(½) f = 48.43494882292° g² = 5+1.93185165258²-2√(5)*1.93185165258cos48.43494882292 g² = 3 g = √(3) units. (√(3)/sin48.43494882292) = (1.93185165258/sinf) f = 56.56505117716° Therefore the required angle alpha is; 180-56.56505117716-atan(2) = 180-120 = 60°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1848404

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Calculating AQ using similar plane shapes (triangle) rule. 12 = 6   6 = a cross multiply. 12a = 36 a = 3 cm. Where a is AQ. PQ = √(6²+3²) PQ =√(45) PQ = 3√(5) cm. BQ = 12-3 BQ = 9 cm. Calculating BR using similar plane shapes (triangle) rule. 6 = 9 3 = b cross multiply. 6b = 27 b = ½(9) cm. b = 4.5 cm. Where b is BR. QR = √(9²+4.5²) QR = ½(9√(5)) cm. Calculating angle PQS. tanc = PS/PQ c = atan((4.5√(5))÷(3√(5))) c = atan(3/2)° Where c is angle PQS. Angle AQP = atan(2)° Angle AQD =...
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