Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1925591

Let BD be 1 unit. Therefore; DC = 2 units. a = 180-60 a = 120° b = 180-120-45 b = 15° (1/sin15) = (c/sin120) c = 3.34606521495 units. d² = 3.34606521495²+3²-2*3.34606521495*3cos45 d = 2.44948974278 units. (2.44948974278/sin45) = (3.34606521495/sine) Where e is the required angle, C. Angle C = 75°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1925426

a² = 2*12² a = 12√(2) cm. Where a is AC = CF. b = 90+45 b = 135° Where b is angle BCE. c² = (12√(2))²+12²-24*12√(2)cos135 c = 12√(5) cm. Where c is BE. (12√(5)/sin135) = (12/sind) d = 18.43494882292° Where d is angle EBC. e = 180-18.43494882292-45 e = 116.56505117708° Where e is angle EGC. (12√(2)/sin116.56505117708) = (f/sin18.43494882292) f = 6 cm. Where f is CG = DG. g = 180-116.56505117708 g = 63.43494882292° Where g is angle CGH. h = 180-63.43494882292-45 h = 71.56505117708= Where...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1924236

Let the single side length of the two congruent inscribed regular pentagon be 1 unit. Area Green is; 0.5*5(1/(2tan(180/5))) = 1.72047740059 square units. Calculating Area Shaded. a = ⅕(180*3) = 108° Where a is the single interior angle of the congruent inscribed regular pentagon. b² = 2-2cos108 b = 1.61803398875 units. c = 1+b c = 2.61803398875 units. d² = 2.61803398875²+1²-2*1*2.61803398875cos72 d = 2.49721204096 units. Where d is the radius of the inscribed arc. 2.49721204096² = 1...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1924072

(1) 2x²+5x+3 = 0 Using smart completing the square approach. (x+(5/4))²+(3/2)-(5/4)² = 0 (x+(5/4))² = (25/16)-(3/2) (x+(5/4))² = (1/16) x = -(5/4)±(1/4) x = -1 Or x = -1.5 Using factorization approach. 2x²+5x+3 = 0 2x²+2x+3x+3 = 0 2x(x+1)+3(x+1) = 0 (x+1)(2x+3) = 0 Therefore; x = -1 Or x = -1.5 (2). 2 = 5 x = 10 Where x is the circumference of the circle. Cross Multiply. 5x = 20 x = 4 cm.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1924012

a² = 3²+4² a = 5cm. Where a is BC. b² = 2.5²+4²-8*2.5cos(atan(3/4)) b = 2.5 cm. c = 90-atan(3/4) c = atan(4/3) sin(atan(4/3)) = d/4 d = 3.2cm. Where d is CN. e = 5-3.2 e = 1.8cm. Where e is BN. Therefore; BN÷CN is; 1.8÷3.2 = 18÷32 = 9/16

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1923083

Let the side of the regular hexagon which is equal the side of the square be 1 unit. a = ½(180-108) a = 36° tan36 = b/0.5 b = 0.363271264 unit. Area Blue is; 0.5*0.363271264*0.5 = 0.090817816 square units. c² = 2-2cos120 c = 1.73205080757 units. d = c-1 d = 0.73205080757 unit. sin54 = e/f f = 1.2360679775e It implies; f+e = d 1.2360679775e+e = 0.73205080757 e = 0.32738307374 unit. Where e is the radius of the inscribed circle. Area Lilac (inscribed circle) is; π(0.32738307374)² =...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1923070

Let GD (inscribed square side) be a. Calculating a. 2(a/(tan60))+a = 12 a = 12/((2/√(3))+1) a = 5.56921938165 cm. b = ½(12-5.56921938165) b = 3.21539030918 cm. Where b is BD = CE. c = 3.21539030918+5.56921938165 c = 8.78460969083 cm. Where c is CD. d² = 3.21539030918²+12²-24*3.21539030918cos60 d = 10.75890566601 cm. Where d is AD. (10.75890566601/sin60) = (12/sine) e = 105° Where e is angle ADB. f = 105-90 f = 15° Where f is angle ADG. cos15 = 5.56921938165/g g = 5.7656801693 cm. Where...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1923013

tana = (4/3) a = atan(4/3)° b = (135-atan(4/3))° (c/sin45) = (6/(135-atan(4/3))) c = 4.28571428571 units. Therefore Area Blue is; 0.5*4.28571428571*6sin(atan(4/3)) = 10.28571428571 square units. = (72/7) square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1922578

Notice; The bigger square side is 1 unit. Calculating the big square side. Let it be a. a² = 2(½)² a = (1/√(2)) units. Calculating the biggest square side. Let it be b. b² = (0.5)²+(1.5)² b = ½√(10) units. tanc = ((0.5)/(1.5)) c = atan(0.5/1.5)° d = (45-atan(0.5/1.5))° e = 90-(45-atan(0.5/1.5)) e= (45+atan(0.5/1.5))° tan(45+atan(0.5/1.5)) = d/(1/√(2)) d = √(2) units. Therefore shaded region is; ½*a*d = ½*(1/√(2))*√(2) = ½ square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1921900

Let the regular pentagon side length be 1 unit. Single interior angle of the regular pentagon is; ⅕(180*3) = 108° a = 108-90 a = 18° b = 180-18-108 b = 54° (1/sin54) = c/sin108) c = 1.17557050458 units. d = 108-36 d = 72° tan72 =e/0.5 e = 1.53884176859 units. f = 54-18 f = 36° g² = 1.53884176859²+1.17557050458²-2*1.17557050458*1.53884176859cos36 g = 0.9071653746 units. (0.9071653746/sin36) = (1.17557050458/sinh) h = 49.61382244038° j² = 1+1.53884176859²-2*1.53884176859cos54 j = 1.248...

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