Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1861301

Let AB = x Let r be the radius of the semi circle. It implies; OC = x-r y = √(r²-(x-r)²) y = √(r²-(x²-2xr+r²)) y = √(2xr-x²) Therefore; 6²+r² = (x-r)²+(2+√(2xr-x²))² 6²+r² = x²-2xr+r²+4+4√(2xr-x²)+2xr-x² 6²+r² = r²+4+4√(2xr-x²) 32 = 4√(2xr-x²) 8 = √(2xr-x²) 64 = 2xr-x² 64+x² = 2xr r = (64+x²)/(2x) -------- (1) z = √((x+r)²-r²) z = √(x²+2xr+r²-r²) z = √(x²+2xr) It implies; (√(x²+2xr)+6)² = (2x)²+(2+√(2xr-x²))² x²+2xr+12√(x²+2xr)+36 = 4x²+4+4√(2xr-x²)+2xr-x² 2x²+12√(x²+2xr)+32 = 4x²+4√(...

Mathematics Question and Solution

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r² = x²+√(8)² x = √(r²-8) units. Calculating r, radius of the semi circle. 3² = x²+(√(8)-(r/2))² 9 = √(r²-8)²+8-2√(2)r+(r²/4) 9 = r²-8+8-2√(2)r+(r²/4) 36 = 4r²-8√(2)r+r² 5r²-8√(2)r-36 = 0 Solving the quadratic equation using completing the square approach. r²-((8√(2)r)/5)-(36/5) = 0 (r-(4√(2)/5))² = (36/5)+(4√(2)/5)² (r-(4√(2)/5))² = (36/5)+(32/25) (r-(4√(2)/5))² = (180+32)/25 r-(4√(2)/5) = ±√(212/25) r = (4√(2)/5)±(2√(53)/5) r = ⅕(4√(2)±2√(53)) r ≠ ⅕(4√(2)-2√(53)) units. r = ⅕(4√(2)+2√(5...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1859501

Please, move a step right to review the analyzed figure. Considering similar triangle ABG ratios. AG = x 12 = 12-x It implies; AG = 12x/(12-x) Notice; Angle ABG = Angle EFI, therefore considering similar triangle ratios of the both triangles. (6-x) = 12x/(12-x) (12-x) = 12 It implies; 12(6-x) = (12-x)(12x/(12-x)) 72-12x = 12x 24x = 72 x = 3 cm. Therefore; AG = 4 cm. DG = 8 cm. Angle ABG = Angle EFI = atan(⅓)° (CG)² = 12²+8² CG = 4√(13) cm. Angle FEI = atan(3)° Angle CEH = (180-...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1859384

Let the forward diagonal of the quadrilateral be 2 units. (2/sin110) = (a/sin35) a = 1.22077458876 units. (b/sin30) = (2/sin55) b = 1.22077458876 units. It implies; x = ½(180-130) x = 25°

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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1858604

Let the base of the ascribed triangle be 1 unit. a = 180-3-6 a = 171° (1/sin171) = (b/sin3) b = 0.33455515209 units. c = 180-30-87 c = 63° (1/sin63) = (d/sin30) d = 0.56116311882 units. e² = 0.56116311882²+0.33455515209²-2*0.33455515209*0.56116311882cos81 e = 0.60670683106 units. Calculating angle x. (0.60670683106/sin81) = (0.33455515209/sinx) x = 33.00000000025° x = 33°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1855059

Calculating BC. (BC/sin30) = (12/sin75) BC = 6.21165708246 cm. Calculating EF. (6.21165708246/sin75) = (EF/sin30) EF = 3.21539030917 cm. Angle FEG is; 360-2(75)-60 = 360-210 = 150° It implies; Area triangle EFG is; 0.5*3.21539030917*3.21539030917sin150 = 2.58468371008 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1855044

Let the radius of the circle be r. Calculating r. √(r²-4)+√(r²-1) = 3 r²-4 = (3-√(r²-1))² r²-4 = 9-6√(r²-1)+r²-1 6√(r²-1) = 12 √(r²-1) = 2 r = √(5) units.

Mathematics Question and Solution

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

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...

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