Mathematics Question and Solution

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

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

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

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

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...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1877416

Let the base of the quadrilateral be 1 unit. (1/sin80) = (a/sin60) a = 0.87938524157 units. (0.87938524157/sin80) = (b/sin30) b = 0.44647558819 units. c² = 0.44647558819²+1-2*0.44647558819cos110 c = 1.2266815969 units. Calculating x, the required angle. (1.2266815969/sin110) = (1/sinx) sinx = 0.76604444312 x = asin(0.76604444312 ) x = 50°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1877260

AD = √(12²+6²) AD = 6√(5) cm. Angle AMD = atan(2√(5))° Angle EBM = (90-ata(2√(5)))° sin(90-ata(2√(5))) = EM/9 EM = 1.96396101212 cm. Angle CME = 180-atan(2√(5)) = 102.60438264838° (CE)² = 9+1.96396101212²-6*1.96396101212cos102.60438264838 CE = 3.92792202425 cm. (3.92792202425/sin102.60438264838) = (1.96396101212/sina) a = 29.20593224736° Where a is angle ECM. cos29.20593224736 = 6/CF CF = 6.87386354243 cm. EF = CF-CE EF = 6.87386354243-3.92792202425 EF = 2.94594151818 cm. Therefore;...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1876454

Let a be the side of the inscribed square. Calculating a. tan60 = a/b b = a/(tan60) cm. Where b is BS = CR. Therefore; 2b+a = 10 2a/(tan60)+a = 10 a((2/tan60)+1) = 10 a = 10(2√(3)-3) cm. a = 4.64101615138 cm. c = ½(10-a) c = 0.5(10-4.64101615138) c = 2.67949192431 cm. Where c = b. d = c+a d = 2.67949192431+4.64101615138 d = 7.32050807569 cm. Where d is BR. e = atan(4.64101615138/7.32050807569) e = 32.37365912705° It implies; Calculating TI, let it be f. tan32.37365912705 = f/g g = f/...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1875478

a² = 6²-1² a = √(35) units. b = atan(√(35)) cos(atan(√(35))) = c/2 c = ⅓ unit. Where c is CE. d = atan(√(35))-60 d = 20.40593177314° Where d is angle FCE. tan20.40593177314 = e/(1/3) e = 0.12400483955 units. Where e is EF. f = 6-(1/3) f = ⅓(17) units. Where f is AE. Therefore; Length AF is; √((⅓(17))²+(0.12400483955)²) = 5.66802331605 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1873085

EG = DG = ½√(12²+6²) EG = DG = 3√(5) cm. Ang CDE = atan(½) It's implies; (CG)² = 45+144-2*36√(5)cos(atan(½)) CG = 3√(5) cm. CG = 6.7082039325 cm. tan(atan(½)) = a/6 a = 3 cm. b = 12-3 b = 9 cm. Angle BAG = atan(9/6) Angle BAG = atan(3/2)° sin(atan(3/2)) = 9/(AG) AG = 9/sin(atan(3/2)) AG = 3√(13) cm. Therefore; AF = AG-FG AF = 3√(13)-3√(5) AF = 3(√(13)-√(5)) cm. AF = 4.10844989389 cm. AF ÷ CG is; 3(√(13)-√(5))÷3√(5) = (√(13)-√(5))/√(5) = ⅕(√(65)-5) = 0.61245154966 Or AF ÷ CG is;...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1872693

Calculating angle ABD. Let it be a. Using cosine rule. 25 = 36+9-2*18cosa 36cosa = 45-25 a = acos(20/36)° a = acos(5/9)° Calculating x. Using cosine rule. x² = 36+144-2*6*12cos(acos(5/9)) x² = 180-144(5/9) x² = 180-(5*16) x² = 100 x = 10 units.

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