Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2049154

Let the big square side length be 1 unit. tan15 = a/1 a = 0.26794919243 units. b = 1-0.26794919243 b = 0.73205080757 units. sin75 = c/0.73205080757 c = 0.70710678119 units. c is the side length of the small square. cos75 = d/0.73205080757 d = 0.18946869098 units. cos15 = 1/e e = 1.03527618041 units. f = 1.03527618041+0.18946869098+0.70710678119 f = 1.93185165258 units. g² = 2*0.70710678119² g = 1 unit. h = 90+45+15 h = 150° j² = 2-2cos150 j = 1.93185165258 units. Calculating angle...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2047755

Let AB be 2 units. tan(atan½) = a/1 a = ½ unit. b = (atan(⅔))° tan(atan⅔) = c/1 c = ⅔ units. d = (180-atan⅔-atan2)° (e/sin(atan2)) = (1/sin (180-atan⅔-atan2)) e = ¼√(13) unit. It implies; Area Blue is; 2(area triangle with height ⅔ unit and base 1 unit - area triangle with height ¼√(13) unit and base sin(atan⅔) units. = 2(½*⅔*1- ½*¼*√(13)*sin(atan⅔)) = 2(⅓ - ¼) = 2(1/12) = ⅙ square units. Calculating Area Shaded. f² = (½)²+(⅓)² f = ⅙√(13) unit. g = f = ⅙√(13) unit. h² = 2(½)² h =...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2047743

Let the side length of the regular pentagon be 1 unit. a² = 2-2cos108 a = 1.61803398875 units. a is the side length of the square. sin18 = b/1 b = 0.30901699437 units. cos18 = c/1 c = 0.9510565163 units. d = 1.61803398875-0.9510565163 d = 0.66697747245 units. e = 1.61803398875-0.30901699437 e = 1.30901699438 units. It implies, the required angle x° is; tanx = 0.66697747245/1.30901699438 x = 27°

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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2046199

Let AC be 2 units. a = √(2²+2²) a = √(8) units. a = 2√(2) units. Let the side length of the inscribed regular hexagon be b. c² = 2b²-2b²cos120 c = √(3)b units. d = 0.5b units. e = √(2) units. It implies; e+d+c = a √(2)+½(b)+√(3)b = 2√(2) ½(b)+√(3)b = √(2) b+2√(3)b = 2√(2) b = 2√(2)/(2√(3)+1) units. b = (4√(6)-2√(2))/11 units. Again, b is the side length of the inscribed regular hexagon. c = √(3)b c = √(3)(4√(6)-2√(2))/11 c = (12√(2)-2√(6))/11 units (((12√(2)-2√(6))/11)/sin90) = (f/sin...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2046134

a² =10²-6² a = 8 cm. tanb = 3/4 b = atan(3/4)° (3/4) = c/10 c = ½(15) cm. c = 7.5 cm. sin(atan(3/4)) = d/8 d = 4.8 cm. e = 7.5-4.8 e = 2.7 cm. (3/4) = 2.7/f f = 3.6 cm. g² = 2.7²+3.6² g = 4.5 cm. h² = 10²-4.8² h = ⅖√(481) cm. h = 8.77268487978 cm. j = ⅖√(481)-3.6 j = 8.77268487978-3.6 j = 5.17268487978 cm. Or j = ⅖√(481)-(18/5) j = ⅕(2√(481)-18) cm. k = (108-atan(¾))° It implies; Area Red Exactly is; 0.5*⅕(2√(481)-18)*4.5sin(180-atan(3/4)) = 27(√(481)-9)/50 cm² = 6.98312458771 cm²...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2045565

Sir Mike Ambrose is the author of the question. Let alpha be x. tanx = 10/a --- (1) tan2x = a/30 --- (2) From (2) a = 30tan2x --- (3) Sub. (3) in (1) tanx = 10/30tan2x tanxtan2x = ⅓ tanx(2tanx/(1-tan²x)) = ⅓ 2tan²x/(1-tan²x) = ⅓ 6tan²x = 1-tan²x 7tan²x = 1 tan²x = ⅐ x = atan(⅐√(7))° x = 20.70481105464° It implies; From (3) a = 30tan2x, and x = 20.70481105464° a = 30tan(2*20.70481105464) a = 26.45751311065 cm. b = 2x b = 41.40962210927° c = 90-b c = 48.59037789073° d = 90-x d = 69.29...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2045384

Let the three congruent semi circles radius be 2 units. Calculating radius, r of the ascribed quarter circle. Let it be a. 2b² = 2² b = √(2) units. 4²-√(2)² = c c = √(14) units. It implies; a = 2+√(14)+√(2) a (radius of the ascribed quarter circle) = 7.15587094915 units. Area Quarter Circle is; ¼(7.15587094915)²π = 40.21748244675 square units. Area R is; 40.21748244675 - 2(½*√(14)*√(2)+½*2+(180-atan(√(2/14))*4π/360+(135-atan(√(7))*4π/360) = 40.21748244675 - 2(2.64575131106+1+5.56...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2042344

Sir Mike Ambrose is the author of the question. Let the side of the regular pentagon be 1 unit. Area regular pentagon is; ½(5)*(1/(2tan(36))) = 1.72047740059 square units. a = ⅕(180*3) a = 108° (1/sin54) = (b/sin18) b = 0.38196601125 unit. c = 1-b c = 0.61803398875 unit. sin36 = (0.61803398875/d) d = 1.05146222424 units. e = asin(0.61803398875/1.05146222424) e = 36° (f/sin27) = (1.05146222424/sin81) f = 0.48330413656 unit. Le the radius of the circle be r. sin72 = r/0.48330413656 r = 0...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2042029

Let the side length of the regular hexagon be 1 unit. a² = 2-2cos120 a = √(3) units. Notice; AB = AC Calculating y. It implies; ((√(3)-1)/sin75) = (y/sin45) y = (4-2√(3)) units. x = 1-y x = 1-(4-2√(3)) x = (2√(3)-3) units. Therefore; x : y is; (2√(3)-3)÷(4-2√(3)) = ½√(3) = sin60

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Mathematics Question and Solution

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