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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3237960

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Calculating the required angle, let it be x. Let a be the side length of the equilateral triangle. It implies; 6² = a²+10²-2*10*a cosy 36 = a²+100-20acosy 20acosy = a²+64 cosy = (a²+64)/(20a) --- (1). 8² = a²+10²-2*10*acos(60-y) 64 = a²+100-20a(½*cosy+½√(3)siny) 64 = a²+100-10acosy-10√(3)asiny 10acosy+10√(3)asiny = a²+36 --- (2). At (1). cosy = (a²+64)/(20a) Calculating siny. siny = opp/hyp opp = √((20a)²-(a²+64)²) opp = √(400a²-(a⁴+128a²+4096)) opp = √(-a⁴...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3237399

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Calculating Area Circle Green. Let x be the radius of area circle green. a²+2² = x² a = √(x²-4) units. b = a+x b = (x+√(x²-4)) units. Calculating x. (x+√(x²-4)) ~ 2 8 ~ √(x²-4) Cross Multiply. 16 = x√(x²-4)+x²-4 (20-x²)² = x²(x²-4) 400-40x²+x⁴ = x⁴-4x² 400 = 36x² x² = 400/36 x = √(400/36) x = ⅙(20) units. x = 3⅓ units. Again, x is the radius of the green circle. It implies; Area green circle is; πx² = ⅑(100π) square units. = 34.9065850399 square un...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3236929

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Calculating Shaded Area. Shaded blue area is  Area quarter of radius 10 cm - Area sector of radius 10 cm and angle 30° - Area equilateral triangle of single side length 5 cm - Area sector of radius 5 cm and angle 120°. = (¼(100π))-((100π)/12)-(¼(25√(3)))-(⅓(25π)) = 25π - ⅓(50π) - ¼(25√(3))  = (300π-200π-75√(3))/12 = (100π-75√(3))/12 cm² = 15.3546212326 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3236887

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Sir Mike Ambrose is the author of the question. Calculating Area yellow ÷ Area purple. Let the single side length of the regular nonagon be 2 units. Therefore; Area purple is; Area triangle with two equal length 2 units each and angle 140°, the angle they form + Area rectangle with length √(3) units and width 1 unit + Area rectangle with length √(8-8cos140) units and width √(3) units. = ½(4sin140)+(1*√(3))+(√(3)*√(8-8cos140)) = 9.52800747773 square units. Area yellow is;...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3231294

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Let Alpha be x. Calculating tanx. Angle ABD = Angle ACD = y. Let AB = AC = a. Let AD = b. b² = a²+10²-2a*10cosy b² = a²+100-20acosy --- (1). b² = a²+6²-2a*6cosy b² = a²+36-12acosy --- (2). cosy = a/10 --- (3). Equating (1) and (2). a²+100-20acosy = a²+36-12acosy 64 = 8acosy 8 = acosy cosy = 8/a --- (4). Substituting (3) in (4). 8/a = a/10 a² = 80 a = √(80) a = 4√(5) units. cosc = 4√(5)/10 c = acos(⅕(2√(5)))° Calculating alpha (x). 180 = 2acos...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3228162

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Calculating the red shaded area. Notice. 10 cm is the radius of each of the four equal inscribed circles. tana = 10/(4*10) a = atan(¼)° a = 14.0362434679° b = 3*10 b = 30 cm. (10/sin14.0362434679) = (30/sinc) c = 133.313856658° d = 180-a-c d = 180-14.0362434679-133.313856658 d = 32.6498998741° e = 90-d e = 90-32.6498998741 e = 57.3501001259° f = 180-2a f = 180-2*14.0362434679 f = 151.927513064° Therefore, the required red shaded area is; Area rectangle...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3227706

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Calculating Parallelogram ABCD Perimeter. 4² = x²+y²-2xycos60 16 = x²+y²-xy 16+xy = x²+y² --- (1). 6² = x²+y²-2xycos120 36 = x²+y²+xy 36-xy = x²+y² --- (2). Equating (1) and (2). 16+xy = 36-xy 2xy = 20 xy = 10 x = 10/y --- (3). Calculating y. Substituting (3) in (1). 16+(10/y)y = (10/y)²+y² 26 = (100/y²)+y² 26y² = 100+y⁴ y⁴-26y²+100 = 0 (y²-13)² = -100+(-13)² (y²-13)² = 169-100 y² = 13±√(69) It implies; y ≠ √(13-√(69)) units. y = √(13+√(69)) units. y...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3226397

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Sir Mike Ambrose is the author of the question. Calculating Exactly Angle Alpha - Angle Beta. Let Alpha be x. Let Beta be y. (6+6√(2))² = 8²+10²-2*8*10cosa 36+72√(2)+72 = 164-160cosa  160cosa = 164-36-72-72√(2) 160cosa = 56-72√(2) 20cosa = 7-9√(2) cosa = (7-9√(2))/20 a = acos((7-9√(2))/20)° a = 106.642318459° 8² = 10²+(6+6√(2))²-2*10(6+6√(2))cosb 64 = 100+36+72√(2)+72-(120+120√(2))cosb (120+120√(2))cosb = 144+72√(2) cosb = (144+72√(2))/(120+120√(2)) cosb = ⅕(3√(2)) b...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3226014

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Calculating Area Red Inscribed Square. Let x be the side length of the red square. a = (5-x) cm. b = 5-½(x) b = ½(10-x) cm. Therefore; Calculating x. 5² = (5-x)²+(½(10-x))² 25 = 25-10x+x²+¼(100-20x+x²) 10x-x² = ¼(100-20x+x²) 40x-4x² = 100-20x+x² 5x²-60x+100 = 0 x²-12x+20 = 0 x²-10x-2x+20 = 0 x(x-10)-2(x-10) = 0 (x-10)(x-2) = 0 It implies; x ≠ 10 cm. x = 2 cm. Therefore, area red square is; x² = 2² = 4 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3224587

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Sir Mike Ambrose is the author of the question. Calculating Green Area to 3 significant figures. Notice. 8 cm is the side length of the regular nonagon and also the regular bigger inscribed pentagon. a = ⅑*180(9-2) a = 140° a is the single interior angle of the regular nonagon. b = ⅕*180(5-2) b = 108° b is the single interior angle of the both inscribed regular pentagons. c = 108-½(180-108) c = 72° tan72 = d/(0.5x) d = 1.53884176859x cm. d is the height of the big...
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