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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3278891

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Calculating angle x. Notice. AC = BC. Let AB be 1 unit. (1/sin(180-70-60-20)) = (a/sin(60+20)) a = 1.96961550602 units. a is AE. (1/sin(180-70-10-60)) = (b/sin60) b = 1.34729635533 units. b is AD. It implies; c² = 1.34729635533²+1.96961550602²-2*1.34729635533*1.96961550602cos10 c = 0.68404028665 units. c is DE. Therefore, angle x is; (0.68404028665/sin10) = (1.34729635533/sinx) x = asin(0.34202014333) x = 20°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3277868

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Calculating the area of the ascribed regular octagon. Let the equal longest length of the both inscribed yellow triangles be x. a = ⅛*180(8-2)) a = ⅛(180*6) a = ½(45*6) a = 135° a is the single interior angle of the ascribed regular octagon. sin(0.5*135) = (0.5x)/b b = 0.54119610015x units. b is the side length of the regular octagon. c = 135-(0.5*45)-½(360-2(135)) c = 135-22.5-45 c = 135-67.5 c = 67.5° c is the second bigger interior angle of inscribed triangle 21 squar...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3277864

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Calculating Shaded Area. a = 2+3 a = 5 units. b = 4+3 b = 7 units. tanc = 7/3 c = atan(7/3)° d² = 2(2)² d = 2√(2) units. tane = 5/(3-2) e = atan(5)° f = 180-e f = (180-atan(5))° tang = 3/7 g = atan(3/7)° h = 180-f-g h = 180-(180-atan(5))-atan(3/7) h = (atan(5)-atan(3/7))° h = 55.4914770123° j = 180-45-atan(7/3) j = (45+atan(3/7))° j = 68.1985905136° k = 180-(45+atan(3/7))-(atan(7)-atan(3/7)) k = 180-68.1985905136-55.4914770123 k = 56.3099324741°...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3276666

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Calculating radius of the circle. Let x be the diameter of the circle. a²+x² = (3+1)² a = √(16-x²) units. It implies; √(16-x²) ~ ½(x) (3+1) ~ 2√(3) Cross Multiply. 2√(48-3x²) = 2x 48-3x²= x² 48 = 4x² x² = 12 x = 2√(3) units. Again, x is the diameter of the circle. Therefore, r, radius of the circle is; ½(x) = ½(2√(3)) = √(3) units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3276554

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Calculating Triangle ADE.                                                          a² = (3√(3))²+(4√(3))².   a² = 27+48  a = √(75)   a = 5√(3) units. a is AC.       3√(3)b+4√(3)b+5√(3)b = 3√(3)*4√(3)  12√(3)b = 36.  √(3)b = 3 b = √(3) units. b is the radius of the inscribed circle. c = 3√(3)-b c = 2√(3) units. d = a-c d = 5√(3)-2√(3) d = 3√(3) units. tane = 3√(3)/√(3) e = atan(3)° f = ½(180-atan(3)) f = 54.2174744115° f is half angle CDE. g = ½(270-atan(4/3)-...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3276512

Calculating Triangle ADE.                                                          a² = (3√(3))²+(4√(3))².   a² = 27+48  a = √(75)   a = 5√(3) units. a is AC.       3√(3)b+4√(3)b+5√(3)b = 3√(3)*4√(3)  12√(3)b = 36.  √(3)b = 3 b = √(3) units. b is the radius of the inscribed circle. c = 3√(3)-b c = 2√(3) units. d = a-c d = 5√(3)-2√(3) d = 3√(3) units. tane = 3√(3)/√(3) e = atan(3)° f = ½(180-atan(3)) f = 54.2174744115° f is half angle CDE. g = ½(270-atan(4/3)-(180-atan(3))) g = 54.2174744...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3275459

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Calculating ratio Red length to Blue length. a² = 2(1)² a = √(2) units. a is the side length of the bigger square. tanb = 1/(1+1) b = atan(½)° c = 45-b c = (45-atan(½))° tan(45-atan(½)) = d/√(2) d = ⅓√(2) units. d is the blue length. e = √(2)-d e = √(2)-⅓√(2) e = ⅓(2√(2)) units. e is the red length. Therefore; Red Length : Blue Length is; ⅓(2√(2)) : ⅓√(2) = 2 : 1

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3275453

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Calculating Area ABCD (square). Notice. The green triangle is an isosceles right-angled triangle. Let the side length of the square be x. Calculating x. (x/sin45) = (a/sin(180-45-30)) a = ½(1+√(3))x units. cos60 = b/x b = ½(x) units  It implies; a+b = 2+√(3) ½(1+√(3))x+½(x) = 2+√(3) (2+√(3))x = 2(2+√(3)) x = 2 units. Again, x is the side length of the square. Area ABCD (square) is; x² = 2² = 4 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3274150

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Calculating The Required Shaded Area. Let it be x. a² = 1²+0.5² a = √(1.25) units. b = 2a b = 2√(1.25) units. tanc = 1/0.5 c = atan(2)° tan(atan(2)) = d/√(1.25) 2 = d/√(1.25) d = 2√(1.25) units. (e/sin45) = (2√(1.25)/sin(180-45-atan(0.5))) e = 1.79505493571 units. f = e-1 f = 0.79505493571 units. It implies, shaded area is; Area trapezoid with parallel lengths 0.79505493571 units and 1 unit respectively, and height 1 unit - Area triangle with height 0.5 units...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3274143

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Calculating Length x. a² = 2²+(2+1)² a = √(13) units. a is the each of the two adjacent lengths of the green isosceles right-angled triangle. tanb = 3/2 b = atan(3/2)° c = 45-atan(2/3) c = 11.309932474° 2√(13)² = d² d = √(26) units. e² = √(26)²+2²-2*2√(26)cos11.309932474 e = 3.16227766017 units. f = 90-atan(3/2) f = 33.690067526° g² = 1+√(13)²-2√(13)cos33.690067526 g = 2.82842712475 units. h² = 2(1)² h = √(2) units. 2.82842712475² = √(2)²+3.16227766017²-2*3...
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