Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1950719

Let the inscribed square side be a. Radius of the half circle is 5 units. Considering Similar Triangle Ratios. 5 = a a = (5-a) Cross Multiply. 25-5a = a² a²+5a-25 = 0 (a+(5/2))² = 25+(5/2)² a+(5/2) = √(25+(25/4)) a+(5/2) = √(125/4) a = -(5/2)±½(5√(5)) a ≠ -½(5+5√(5)) units. a = ½(5√(5)-5) units. Therefore; Area of the inscribed square is; a² = (½(5√(5)-5))² = ¼(125-50√(5)+25) = ¼(150-50√(5)) square units. = 12.5(3-√(5)) square units. = 9.54915028125 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1949823

Let AB be 1 unit. (1/sin100) = (b/sin30) b = 0.50771330594 units. (1/sin110) = (d/sin30) d = 0.53208888624 units. d is CB. (0.50771330594/sin30) = (c/sin50) c = 0.77786191343 units. a² = 0.77786191343²+0.53208888624²-2*0.77786191343*0.53208888624cos10 a = 0.27014860749 units. Therefore, showing that; c = a+b Again; a = 0.27014860749 units. b = 0.50771330594 units. c = 0.77786191343 units. It implies; a+ b = 0.27014860749+0.50771330594 = 0.77786191343 units, which is c. Showed!

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1949559

Notice; The inscribed star is formed from a pentagram (a regular pentagon is ascribing the star). Let the side of the pentagram be 1 unit. tan54 = a/0.5 a = 0.68819096024 unit. cos54 = 0.5/b b = 0.85065080835 unit. c = b+a c = 1.53884176859 units. Let the side length of the ascribed regular hexagon be d. Calculating d. sin45 = e/d e =0.5√(2)d unit. It implies; e+d = c 0.5√(2)d+d = 1.53884176859 d = 0.90143263769 unit. f = 0.90143263769+2(0.90143263769sin45) f = 2.17625089948 units...

Mathematics Question and Solution

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Notice; The figure is not drawn to scale. Let BC be 1 unit. (1/sin135) = (a/sin30) a = 0.70710678119 units. (1/sin135) = (b/sin15) b = 0.36602540378 units. Notice; MB = NC, let it be c. Therefore; MN = (1-2c) units. (AM)² = 0.36602540378²+c²-2*0.36602540378ccos30 (AM)² = 0.13397459621+c²-0.63397459621c --- (1) (AN)² = 0.70710678119²+c²-2*0.70710678119ccos15 (AN)² = 0.5+c²-1.36602540379c --- (2) Therefore; Calculating c. (MN)² = (AM)²+(AN)² (1-2c)² = 0.13...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1948290

Let the inscribed purple regular heptagon side be 1 unit. a = ⅐(180*5) a = ⅐(900)° b = 180-⅐(900)) b = ⅐(360)° c = 180-2(⅐(360)) c = ⅐(540)° (1/sin⅐(540)) = (d/sin⅐(360)) d = 0.8019377358 unit. e = 2(0.8019377358)+1 e = 2.60387547161 units. f = 90-½(⅐(540)) f = ⅐(360)° cos⅐(360) = g/2.60387547161 g = 1.62348980186 units. h = 2g h = 3.24697960372 units. Where h is the base of the ascribed quadrilateral. cos(180/7) = i/0.8019377358 i = 0.72252093395 unit. j = 2i j = 1.4450418679 units...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1948278

Let the congruent lengths be 1 unit. (1/sin50) = (a/sin80) a = 1.28557521937 units. sin(60) = b/2 b = √(3) units. b = 1.73205080757 units. c = 20° d² = 1.73205080757²+1.28557521937²-2*1.73205080757*1.28557521937cos20 d = 0.68404028665 units. Therefore, the required angle is; Let it be e. (0.68404028665/sin20) = (1.28557521937/sine) e = 40°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1948081

Let the square side length be 1 unit. Therefore, the regular hexagon side length is 1 unit as well. sin30 = a/1 a = 0.5 units. b = 1+a b = 1+0.5 b = 1.5 units. cos30 = c/1 c = ½√(3) units. d = 1+c d = 1+½√(3) d = ½(2+√(3)) units. e = 3+c e = 3+½√(3) e = ½(6+√(3)) units. Therefore; angle ABC is; atan(d/b) + atan(e/b) = atan(½(2+√(3))/1.5) + atan(½(6+√(3))/1.5) = 51.206023113+68.793976887 = 120°

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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1946655

Let the height and base of area triangle C be 2 units and 3 units respectively. Therefore; Area Triangle C is; ½*2*3 = 3 square units. tana = (2/3) a = atan(2/3)° b = 189-a b = 180-atan(2/3) b = 146.30993247402° c = ½(180-146.30993247402) c = 16.84503376299° Let the radius of the big circle be d. Calculating d. e = ½(146.30993247402) e = 73.15496623701° f = (3-d) units. tan73.15496623701 = d/(3-d) d = 9.9083269132-3.30277563773d 4.30277563773d = 9.9083269132 d = 9.9083269132/4.3027...

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