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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2546058

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tana = 16/8 a = atan(2)° b² = 16²+8² b = 8√(5) cm. Let BC = c. Therefore; c+3c = b 4c = 8√(5) c = 2√(5) units. Again, c is BC. AB = 3c AB = 3*2√(5) AB = 6√(5) cm. d = 180-30-atan(2) d = (150-atan(2))° (8/sin(150-atan(2))) = (e/sin(atan(2))) e = 7.1682956077 cm. sin60 = f/7.1682956077 f = 6.2079260981 cm. h = 16-f h = 16-6.2079260981 h = 9.7920739019 cm. h is length b. cos(atan(2)) = 2√(5)/j j = 10 cm. k = 16-j k = 6 cm. k is length a. It implies...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2545814

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Let the square side be a. b = 45-20 b = 25° c² = 2a² c = √(2)a Calculating a. a² = 1²+(√(2)a)²-2*1*√(2)acos25 a² = 1+2a²-2.5634255282a a²-2.5634255282a+1 = 0 It implies; a = 2.08345 units. Again, a is the side length of the square. (1/sind) = (2.08345/sin25) d = 11.703400178° Therefore, the required angle theta is; d+45 = 11.703400178+45 = 56.703400178°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2545796

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Let the diameter of the inscribed half circle be 2 units. It radius is 1 unit. Calculating x, angle alpha. tana = 1/0.5 a = atan(2)° b = 2a b = 2atan(2)° c = 180-b c = (180-2atan(2))° d² = 2(0.5)²-2(0.5)²cos(180-2atan(2)) d = 0.4472135955 units. e = ½(180-(180-2atan(2))) e = ½(2atan(2))° e = atan(2)° f = 90-e f = atan(½)° It implies; g² = 0.4472135955²+2²-2*2*0.4472135955cos(atan(½)) g = 1.6124515497 units. (1.6124515497/sin(atan(0.5))) = (0.4472135955...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2545752

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Notice! The inscribed blue area is a square. Let the inscribed blue area side length be a. Calculating a. tan63 = a/b b = a/tan63 --- (1). tan43 = a/c c = a/tan43 --- (2). It implies; (2)+a+(1) = 10 (a/tan43)+a+(a/tan63) = 10 1.07236871a+a+0.5095254495a = 10 2.5818941595a = 10 a = 3.8731254584 cm. Again, a is the side length of the inscribed blue square (square). Therefore, area blue is; a²  = 3.8731254584² = 15 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2545733

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a = (10-b) cm. tan30 = (10-b)/10 10 = 10√(3)-√(3)b b = (10√(3)-10)/√(3) b = ⅓(30-10√(3)) cm. tan30 = ⅓(30-10√(3))/c 3c = 30√(3)-30 c = (10√(3)-10) cm. d = 10-c d = 10-(10√(3)-10) d = (20-10√(3)) cm. Therefore; Area blue is; ½*d*10 = ½*(20-10√(3))*10 = 50(2-√(3)) cm² = 13.3974596216 cm²

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/2543787

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Calculating r, radius of the circle. a² = 4 a= 2 cm. a is the side length of the small square. b = (r-2) cm. It implies; r² = 4²+(r-2)² r² = 16+r²-4r+4 4r = 20 r = 5 cm. Again, r is the radius of the circle. Calculating x, side length of the big square. c = (x-5) cm. It implies; 5² = (x-5)²+√(21)² 25-21 = x²-10x+25 x²-10x+21 = 0 Resolving the above quadratic equation via factorization approach to get x. x²-7x-3x+21 = 0 x(x-7)-3(x-7) = 0 (x-3)(x-7) = 0...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2543745

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c² = 7²+24² c = 25 units. a² = (7+25)²+24² a = 40 units. b = ½(a) b = ½√(40) units. b = 20 units. d²+20² = 25² d = √(625-400) d = √(225) d = 15 units. It implies, R radius of the circle is; R(15+20+25) = (15*20) 60R = 300 R = 5 units. Calculating x. x² = 20²+5² x = √(425) x = 5√(17) units. x = 20.6155281281 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2542212

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a*15 = 540 a = 108/3 a = 36 cm. a is AB. b = a-R b = (36-R) cm. b is OB. It implies; Calculating R. R² = 15²+(36-R)² R² = 225+1296-72R+R² 72R = 1521 R = 21.125 cm.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2541591

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Calculating the area of the blue inscribed rectangle and the area of the ascribed regular pentagon. πr² = 36π r = 6 cm. r is the radius of the inscribed circle. sin(0.5*108) = 6/a a = 7.416407865 cm. b = r+a b = 13.416407865 units. tan36 = 13.416407865/c c = 18.4661012231 cm. d = 2c d = 36.9322024461 cm. d is the width of the inscribed blue rectangle. Therefore; Area Blue Inscribed Rectangle is; 36.9322024461*50 = 1846.6101223052 cm² ≈ 1850 cm² to 3 significan...
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