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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2551427

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tana = 4/7 a = atan(7/4)° b = 180-a b = (90+atan(4/7))° c² = 7²+4² c = √(49+16) c = √(65) units. Therefore, Blue Area is; Area triangle with height √(65) units and base 4sin(90+atan(4/7)) units. = 0.5*√(65)*4sin(90+atan(4/7)) = 14 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2551415

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a = 180-44 a = 136° b = ½(180+44) b = ½(224) b = 112= c = ½(a) c = 68° d = ½(b) d = 56 e = 180-c-d e = 180-68-56 e = 180-124 e = 56° Where e is the required angle, angle CBO. It implies triangle OBC is isosceles.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2551398

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7² = 5²+6²-2*5*6cosa 60cosa = 25+36-49 cosa = (61-49)/60 a = acos(12/60) a = 78.4630409672° Calculating r, radius of the inscribed half circle. 5r+6r = 5*6sin78.4630409672 11r = 30sin78.4630409672 r = (30sin78.4630409672)/11 r = 2.6721706285 units. Again, r is the radius of half the inscribed circle.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2550654

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tan30 = a/h a = htan30 units. tan60 = b/h b = htan60 units. It implies; a+b = 100 htan30+htan60 = 100 ⅓(√(3)h)+√(3)h = 100 ⅓(4√(3)h) = 100 4√(3)h = 300 h = 300/(4√(3)) h = ¼*⅓*300√(3) h = 25√(3) units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2550595

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a²+R² = 7² a = √(49-R²) units. b = (3+R) units. c²+R² = 3² c = √(9-R²) units. d = a+c d = √(49-R²)+√(9-R²) units. R is; (√(49-R²)+√(9-R²))² = (3+R)²+√(49-R²)² 49-R²+2√((49-R²)(9-R²))+9-R² = 9+6R+R²+49-R² 58-2R²+2√((49-R²)(9-R²)) = 58+6R 2√((49-R²)(9-R²)) = 2R²+6R √((49-R²)(9-R²)) = R²+3R (49-R²)(9-R²) = (R²+3R)² 441-58R²+R⁴ = R⁴+6R³++9R² 441-67R²-6R³ = 0 6R³+67R²-441 = 0 It implies; R = ⅓(7) units. R = 2⅓ units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2549647

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0.5*a*6sin75 = 9 a = 9/(3sin75) a = 3/sin75 a = 3.1058285412 units. a is AB. b² = 3.1058285412²+6²-2*3.1058285412*6cos75 b = 6 units. b is AC. It implies, triangle ABC is isosceles. Therefore, the required angle alpha, let it be c is; c = 180-2*75 c = 180-150 c = 30° Again, c is the required angle, alpha.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2549160

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Calculating length CD. Notice! The composite plane shape is not drawn to scale. tan60 = 8/a a = 8/√(3) a = ⅓(8√(3)) units. a is OE. (⅓(8√(3)))²+8² = b² b² = ⅓(64)+64 b² = ⅓(64+192) b² = ⅓(256) b = ⅓(16√(3)) units. b is CE. Or sin60 = 8/c c = 8/sin60 c = 16/√(3) c = ⅓(16√(3)) units. c is CE. (8/sin120) = (⅓(8√(3))/sind) d = 30° d is angle ODE. It implies that DE = OE = ⅓(8√(3)) units. Calculating CD. Let it be e. e² = (⅓(8√(3)))²+(⅓(16√(3)))²-2*⅓(8√(3))*⅓...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2548386

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Let the square side length be 2 units. Therefore, r, radius of the circle is; r = 1 unit. Calculating the required angle, let it be a. cosb = 1/2 b = acos(1/2) b = 60° c² = 2²+2² c = √(8) c = 2√(2) units. c is a diagonal of the square. d = ½(c) d = √(2) units. e = 45+60 e = 105° It implies; f² = √(2)²+1²-2√(2)cos105 f = 1.9318516526 units. (1.9318516526/sin105) = (1/sinh) h = 30° Therefore, the required angle, angle a is; a = h+45 And h = 30° It im...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2548180

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Notice! The side length of the inscribed biggest square is 5 units. Calculating inscribed red area. tana = 2.5/5 a = atan(½)° b = 2atan(½)° tanc = 5/2.5 c = atan(2)° d = 2atan(2)° Therefore, area red is; Area sector with radius 2.5 units and angle 2atan(2) - Area triangle with height 2.5 units and base 2.5sin(2atan(2)) units + Area sector with radius 5 units and angle 2atan(½) - Area triangle with height 5 units and base 5sin(2atan(½)) units - 2(area quarter circle wit...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2548054

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Let the side length of the three equal regular triangles be 2a units each. b² = 2a²-2a²cos120 b = √(3)a unit. (6√(7))² = (2a)²+(√(3)a)²-2*2a*√(3)acos90 252 = 4a²+3a² 252 = 7a² a² = 36 a = 6 units. It implies; 2a = 2*6 = 12 units. Where 12 units is the side length of the three equal regular triangles each. b = √(3)a And a = 6 units. b = 6√(3) units. Therefore, paint area is; 3(area triangle with height 6√(3) units and base 12 units + area triangle with height 6 u...
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