Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2784404

Sir Mike Ambrose is the author of the question. a² = 10²+16²-2*10*16cos120 a = 2√(129) units. Where a is length AC. (3√(129)/sin120) = (16/sinb) b = 37.58908946897° c = 90-b c = 52.41091053103° d = 60-b d = 22.41091053103° e = 90-d e = 67.58908946897° f = 90-60 f = 30° cos37.58908946897 = g/10 g = 7.92405815693 units. i = 30+37.58908946897 i = 67.58908946897° tan67.58908946897 = j/7.92405815693 j = 19.2148198619 units. Where j is the radius of the arc...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2784373

Let the side length of the regular hexagon be 1 unit. Therefore; a = 1 unit. Calculating b. Cos30=a/c And a = 1 c = 1/cos30 c = ⅔√(3) units. Where c is the hypotenuse of the right-angled triangle formed with length a. It implies; b = √(a²+c²) b = √(1²+(⅔√(3))²) b = √(1+(4/3)) b = √(7/3) units. Therefore; b/a is; √(7/3) ÷ 1 = √(7/3) = ⅓√(21) = 1.52752523165

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2784356

Sir Mike Ambrose is the author of the question. Area Blue, in cm² to 1 d. p. is; Half area regular pentagon with side 12 cm - area triangle with height 6 cm and base 6tan24 cm - area triangle with height 12cos36 and base 12sin36 cm. = (½*(60*12)/(4tan(180/5))) - ½(6*6tan24) - ½(12cos36*12sin36) = 123.874372842 - 8.01411633555 - 34.2380345866 = 81.62222192 cm² ≈ 81.6 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2783541

Calculating yellow area. a = ½(10) a = 5 km. a is the radius of the inscribed circle. tanb = 5/10 b = atan(½)° c = 180-2b c = (180-2atan(½))° d² = 2(5²)-2(5²)cos(180-2atan(½)) d = 8.94427191 km. e² = 5²+5² e = √(50) e = 5√(2) km. f = 45-b f = (45-atan(½))° Therefore, area yellow is; 0.5*5√(2)*8.94427191sin(45-atan(½)) = 10 km²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2783001

Let a be the radius of the inscribed circle. Let b be the radius of the ascribed circle. It implies; πb²-πa² = 20π b²-a² = 20 b = √(20+a²) units. Again, b is the radius of the ascribed circle. c = 2b c = 2√(20+a²) units. c is the ascribed circle's diameter. 2d² = (2√(20+a²))² 2d² = 4(20+a²) d² = 40+2a² d = √(40+2a²) units. d is AB. e = ½(d) units. e = ½√(40+2a²) units. It implies, calculate a. √(20+a²)² = a²+(½√(40+2a²))² 20+a² = a²+¼(40+2a²) 20 = 10+½(a²)...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2782863

Sir Mike Ambrose is the author of the question. Let the side length of the large square be x. Calculating x. 25=x²+64-(16xcos20) Therefore; x = 11.70244 cm. It implies; Area large square is; 11.70244² = 136.947101954 cm² Area orange is; Area triangle with height and base 8 cm each - Area triangle with two side 2.91176187413 cm and 2.27177411013 cm, and angle 70° = ½(8)² - ½(2.91176187413*2.27177411013)sin70 = 32 - 3.107970027 = 28.892029973 cm² Therefore; Are...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2782853

Sir Mike Ambrose is the author of the question. Let the side length of the regular pentagon be 2 units. a² = 2²-1² a = √(3) units. b² = 5-4cos108 b = 2.49721204096 units. (2.49721204096/sin108) = (2/sinc) c = 49.61382244056° d = 90-c d = 40.38617755944° e² = 3+2.49721204096²-2√(3)*2.49721204096cos40.38617755944 e = 1.62694657231 units. (1.62694657231/sin40.38617755944) = (√(3)/sinf) f = 43.61382244075° g = 180-e-f g = 96° Let the radius of the inscribed circle b...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2782725

Calculating Area Red. tana = 5/10 a = atan(½)° b = 2a b = 2atan(½)° c = b-45 c = 8.1301023542° d² =10²+10²-2*10*10cos8.1301023542 d = 1.4177804018 units. e = ½(180-8.1301023542) e = 85.9349488229° f² = 2*5² f = 5√(2) units. g = 10-f g = 2.9289321881 units. tanh = 10/5 h = atan(2)° j = 2h j = 2atan(2)° k = j-90 k = 36.8698976458° It implies, area red is; Area triangle with height 1.4177804018 units and base 2.9289321881sin85.9349488229 units + Area s...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2782384

Let EB be a. Let AE be b. It implies; a/b = √(2) --- (1). a+b = 1 --- (2). From (1). a = √(2)b --- (3). Substituting (3) in (2) to get b. √(2)b+b = 1 b(√(2)+1) = 1 b = (√(2)-1)/(2-1) b = √(2)-1 b = 0.4142135624 units. b is AE. At (2). a+b = 1 And b = 0.4142135624 units. a = 1-0.4142135624 a = 0.5857864376 units. a is EB. tanc = 1/0.4142135624 c = 67.5° c is angle AED. d =½(c) d = 33.75° Let x be the small inscribed circle radius. e = (0.4...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2781334

Let the radius of the circle be 2 units. a = (2-b) units. Calculating b. 2² = 2(2-b)² 4 = 2(4-4b+b²) b²-4b+2 = 0 (b-2)² = -2+(-2)² (b-2)² = 2 b = 2±√(2) It implies; b ≠ (2+√(2)) b = (2-√(2)) units. b = 0.5857864376 units. c = 2-b c = 2-0.5857864376 c = 1.4142135624 units. c = √(2) units. Therefore; The required angle, theta is; Let it be d. d = 45-atan(√(2)/2)) d = 45-35.2643896828 d = 9.7356103172° Therefore; tan(theta) is; tand tan(9.735610317...

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