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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1886366

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Let AB = BC = CD = 1 unit. AC = √(1²+1²) AC = √(2) units. (AD)² = 3-2√(2)cos105 AD = 1.93185165258 units. Therefore; (1.93185165258/sin105) = (√(2)/sin?) ? = 45° Where ? is angle ADC.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1885846

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a² = 4²+4²-2*4*4cos150 a = 7.72740661031 cm. Where a is BE. b = angle ABE = ½(180-150) b = 15° c = angle AFE = 180-45-15 c = 120° (4/sin120) = (d/sin45) d = 3.26598632371 cm. Where d is EF. Notice; EF = CF Therefore; Length CF = 3.26598632371 cm.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1885237

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a = asin(1/6) b = 2a b = 2asin(1/6) Where b is angle BAC cos(2asin(1/6)) = c/6 c = ⅓(17) units. Where c is AD. d = 6-c d = 6-⅓(17) d = ⅓ units Where d is CD. sin(2asin(1/6)) = e/6 e = 1.97202659437 units. Where e is BD. f = asin((1/3)/2) f = 9.59406822686° Where f is angle EBC = angle CAE. tan(9.59406822686) = g/(17/3) g = 0.95784148869 units. Or  g = 17√(35)/105 units. Where g is DE.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1883742

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Notice; Since angle BDC equals angle BAC which are the opposite inscribed angles of quadrilateral ABCD. And; CD+BD = 40 units. It implies; CD+BD = AC+AB It also implies that; Area Triangle BDC = Area Triangle BAC Therefore; AC+AB = 40 units. AC = AB = ½(40) AC = AB = 20 units. Area quadrilateral ABCD is; 2(½*20*20) = 400 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1882433

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9 = 8-8cosa 8cosa = -1 a = acos(-1/8) a = 97.18075578146° Therefore; a• = 180-a a• = 180-97.18075578146  a• = 82.81924421854° b = ½(180-82.81924421854) b = 48.59037789073° c = ½(48.59037789073) c = 24.29518894536° Where c is alpha. sin24.29518894536 = d/4 d = 1.64575131106 cm. cos24.29518894536 = e/4 e = 3.64575131106 cm. f = 90-24.29518894536 f = 65.70481105464° Red area is; 0.5*1*1.64575131106sin65.70481105464 = ¾ cm² Blue area is; 2(0.5*3.64575131106*1.64575131106)-(¾) = 6-(3/4)...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1881678

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a = acos(3/5) Where a is angle GCA. tan(acos(3/5)) = b/3 b = 4 units. Where b is AG. c = atan(⅔) Where c is angle GBE. d = 180-acos(3/5)-atan(⅔) d = 93.17983011986° Where d is angle BEC. (6/sin93.17983011986 ) = (e/sin(acos(⅗))) e = 4.80740170062 units. Where e is BE. Area Parallelogram BCDE is; 2(area triangle with height 4.80740170062 units and base 6sin(atan(⅔)) units. = 2(0.5*4.80740170062*6sin(atan(⅔))) = 16 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1881048

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Let the inscribed square side be a. tan60 = a/b a = √(3)b cm. Where b is BS = CR It implies; b+√(3)b+b = 10 (2+√(3))b = 10 b = 10(2-√(3)) cm. Therefore; a = √(3)*10(2-√(3)) a = 10(2√(3)-3) cm. a = 4.64101615138 cm. sin60 = 10(2√(3)-3)/c c = 20(2-√(3)) cm. Where c is CQ. d² = 10²+(10(2-√(3)))²-20*10(2-√(3))cos60 d = 8.96575472168 cm. Where d is AS. (8.96575472168/sin60) = (10(2-√(3))/sine) e = 15° Where e is angle PAS. f = 180-60-15 f = 105° Where f is angle BSA. g = 105-90 g = 15° Whe...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1880138

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xy = 40 ------ (1).   b = 4+x-5 b = x-1 y(x-1) = 20 ------ (2). At (1) x = 40/y ------ (3). Notice; y is the length or base of the rectangle. Substituting (3) in (2) to get y. y((40/y)-1) = 20 40-y = 20 y = 40-20 y = 20 cm.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1879482

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Making a circle to ascribe A, B, C and E quadrilateral with the circumference of the circle cutting across A, B, C and E. Therefore; Angle ABC will be; ½(240) = 120° Parallelogram ABCD height is; Let it be x. sin60 = x/4 ½√(3) = x/4 x= 2√(3) cm. Parallelogram ABCD Area is; 6*2√(3) = 12√(3) cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1879295

mathematics-question-and-solution_16864104766pxmz2.jpeg
Let the inscribed square side be a. tan60 = a/b a = √(3)b cm. Where b is BS = CR It implies; b+√(3)b+b = 10 (2+√(3))b = 10 b = 10(2-√(3)) cm. Therefore; a = √(3)*10(2-√(3)) a = 10(2√(3)-3) cm. sin60 = 10(2√(3)-3)/c c = 20(2-√(3)) cm. Where c is CQ. d² = 10²+(10(2-√(3)))²-20*10(2-√(3))cos60 d = 8.96575472168 cm. Where d is AS. (8.96575472168/sin60) = (10(2-√(3))/sine) e = 15° Where e is angle PAS. f = 180-60-15 f = 105° Where f is angle BSA. g = 105-90 g = 15° Where g is angle PSA. ta...
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