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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2546667

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a²+8² = 10² a = 6 units. b = ½(10) b = 5 units. b is the radius of the ascribed semi circle. 6² = 5²+5²-2*5*5cosc 50cosc = 50-36 cosc = 14/50 c = acos(7/25)° c = 73.7397952917° d = 2c d = 147.4795905834° e² = 5²+5²-2*5*5cos147.4795905834 e = 9.6 units. f = ½(e) f = 4.8 units. cosg = 4.8/6 g = acos((24/5)/6) g = acos(⅘)° g = 36.8698976458° h = ½(g) h = ½(acos(⅘))° h = 18.4349488229° tan(½(acos(⅘))) = j/(4.8) j = ⅓*(24/5) j = ⅕(8) units. Where j is the...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2545391

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sin45 = 12/a a = 12√(2) cm. a is AC. Notice ! AB = 2BC Let BC = b Therefore; AB = 2b cm. AC = AB+BC AC = 2b+b AC = 3b cm. And AC = 12√(2) cm. It implies; 12√(2) = 3b b = 4√(2) cm. Again, b is BC. AB = 2b  And b is 4√(2) cm. AB = 2*4√(2) AB = 8√(2) cm. c = 180-30-45 c = 105° (12/sin105) = (d/sin30) d = (6/sin105) cm. d = 6.2116570825 cm. e = 8√(2)-d e = (8√(2)sin105-6)/sin105 cm. e = 5.1020514165 cm  (((8√(2)sin105-6)/sin105) /sin30) = (f/sin1...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2540051

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Let AD = CD = a. Calculating a. b² = 2a² b = √(2)a units. b is AC. c²+3² = a² c = √(a²-9) units. c is DE. a² = d²+7² d = √(a²-49) units. e = c-d e = (√(a²-9)-√(a²-49)) units. e is CB. It implies; b² = e²+10² (√(2)a)² = (√(a²-9)-√(a²-49))²+100 2a² = a²-9-2√((a²-9)(a²-49))+a²-49+100 2√((a²-9)(a²-49)) = 42 √((a²-9)(a²-49)) = 21 (a²-9)(a²-49) = 441 a⁴-49a²-9a²+441 = 441 a⁴-58a² = 0 a² = 58 a = √(58) units. Therefore, Orange Area is; ½(a²) = ½*√(58)² =...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2536518

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Let the two equal lengths be a  a²+x² = 10² a = √(100-x²) units. b = ½(8+10) b = 9 units. c²+(9-8)² = x² c = √(x²-1) units. Observing similar plane shape (right-angled) side length ratios. 10 - √(100-x²)   x - √(x²-1) Cross Multiply. 10√(x²-1) = x√(100-x²) 100(x²-1) = x²(100-x²) 100x²-100 = 100x²-x⁴ x⁴ = 100 x² = 10 x = √(10) units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2532052

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Calculating the radius of the small inscribed pink circle. Let it be r. b = ⅕*180(5-2) b = 108° b is the single interior angle of the ascribed regular pentagon. c² = 10²+10²-2*10*10cos108 c = 16.1803398875 units. d = 108-36 d = 72° e = 90-72 e = 8° f²+(0.5*10)² = c² f² = 16.1803398875²-5² f = 15.3884176859 units. tan54 = g/5 g = 6.8819096024 units. Again, a is the radius of the bigger inscribed circle. h = 2g h = 13.7638192047 units. It implies; sin18= r/j...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2526144

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Calculating x. a²+(x-3)² = 10² a² = 100-x²+6x-9 a² = 91+6x-x² a = √(91+6x-x²) cm. It implies; (a+x)² = (x-3)²+(x+4)² (√(91+6x-x²)+x)² = x²-6x+9+x²+8x+16 91+6x-x²+2x√(91+6x-x²)+x² = 2x²+2x+25 2x√(91+6x-x²) = 2x²-4x-66 x√(91+6x-x²) = x²-2x-33 x²(91+6x-x²) = (x²-2x-33)² 91x²+6x³-x⁴ =x⁴-2x³-33x²-2x³+4x²+66x-33x²+66x+1089 91x²+6x³-x⁴ = x⁴-4x³-62x²+132x+1089 2x⁴-10x³-153x²+132x+1089 = 0 Resolving the above fourth power equation. It implies; x = 11 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2520273

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a² = 6²+6²-2*6*6cos120 a = 6√(3) units. tanb = 3/(6√(3)) b = atan(2/(2√(3))) b = 16.102113752° c = 90-16.102113752 c = 73.897886248° d = 30+16.102113752 d = 46.102113752° It implies;  cos46.102113752 = e/6 e = 4.1602514717 units. Therefore, area blue is; 0.5*4.1602514717*6sin73.897886248 = 11.9911209755 square units. ≈ 12 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2512953

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a = ⅐*180(7-2) a = ⅐(900)° a is the single interior angle of the regular heptagon. (√(2)/sin(900/7)) = (1/sinb) b = 33.5619539118° c = 180-(900/7)-b c = 17.8666175168° (d/sin17.8666175168) = (√(2)/sin(900/7)) d = 0.5549581321 units. e = 1-d e = 0.4450418679 units. f² = 2-2cos(900/7) f = 1.8019377358 units. g = ½(180-(900/7)) g = 25.7142857143° h = (900/7)-g h = 102.8571428571° Therefore, length AB is; Let it be j. j² = 1.8019377358²+0.4450418679²-2*1.80193...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2508440

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Considering the quadrilateral cyclic. a² = 2*10²-2*10²cosx a² = 200-200cosx --- (1). a² = 9²+7²-2*9*7cosx a² = 130+126cosx --- (2). Equating (1) and (2). a² = a² 200-200cosx = 130+126cosx 200-130 = 126cosx+200cosx 326cosx = 70 x = acos(70/326) x = 77.6006688181° y = 180-x y = 102.3993311819° Therefore, area cyclic quadrilateral is; Area triangle with height 10 units and base 10sin77.6006688181 units + Area triangle with height 9 units and base 7sin102.3993311819...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2506798

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Let the base of the ascribed triangle be 1 unit. a = 180-12-42-12-36 a = 180-102 a = 78° (1/sin(12+36)) = (b/sin78) b = 1.3162274259 units. c = 180-12-42 c = 180-54 c = 126° (1.3162274259/sin126) = (d/sin12) d = 0.3382612127 units. f² = 0.3382612127²+1-2*0.3382612127cos12 f = 0.6728163648 units. It implies, the required angle x, is; (0.6728163648/sin12) = (0.3382612127/sinx) x = 6.0442188698°
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