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.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1872609

BC = √(√(3)²+1²) BC = 2 units. Let r be the radius of the circle. Using similar triangles ratio rule on triangle ACE and triangle BCE to calculate r. 1 = √(3) √(3) = AE Cross Multiply. AE= 3 units. Therefore; r = ½(AE+BE) r =½(3+1) r = 2 units. Angle ADC is; 180-atan(√(3)) = 180-60 = 120° Therefore shaded area or area triangle ACD is; ½*2*2sin120 = 2sin120 = √(3) square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1871923

Let the square side be a. tan60 = a/b b = a/tan60 cm. Where b is BS. Calculating a. 2b+a = 6 2(a/tan60)+a = 6 a((2/tan60)+1) = 6 a = 6/((2/tan60)+1) a = 6(2√(3)-3) cm. Area Shaded STR is; ½(a²) = ½(6(2√(3)-3))² = 54(7-4√(3)) cm² = 3.87702556513 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1870080

Notice; Diameter of the ascribed semi circle is 20 cm. Radius, r will be 10 cm. Calculating Area Shaded. It will be; Area half circle with radius 10 cm - 2(area semi circle with radius 5 cm) - Area sector with radius 5 cm and angle 60° + 2(area sector with radius 5 cm and angle 60°) - 2(area triangle with two equal lengths 5 cm each and the angle they form, 60°) = ½(100π)-2(½*25π)-⅙(25π)+2(⅙*25π)-2(½*25*(½√(3))) = 50π-25π-⅙(25π)+⅓(25π)-½(25√(3)) = 25π-⅙(25π)+⅓(25π)-½(25√(3...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1868239

Let the side length of the square be 2 units. Let the radius of the small inscribed circle be x. Let the radius of the big inscribed circle be y. Calculating y. (2-y)² = y²+y² 2-y = √(2)y 2 = √(2)y+y It implies; y = 2/(√(2)+1) units. y = 0.82842712475 units Where y is EG. Calculating x. (2+x)² = 2(2-x)² 4+4x+x² = 2(4-4x+x²) 4+4x+x² = 8-8x+2x² x²-12x+4 = 0 Solving the quadratic equation using completing the square approach to get x. (x-6)² = -4+(-6)² (x-6)² = 32 x-6 = ±4√(2) x = 6±4√(2)...

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