Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2515207

a = 180-60 a = 120° b² = 4²+6²-2*4*6cos120 b = 2√(19) units. b = 8.7177978871 units. b is the side length of the inscribed regular triangle ABC (equilateral triangle). (8.7177978871/sin120) = (4/sinc) c = 23.4132244464° d = 60+c d = 83.4132244464° Therefore, length BD is; (BD)² = 6²+8.7177978871²-2*6*8.7177978871cos83.4132244464 BD = 10 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2515184

a² = 3²+4² a = √(25) a = 5 units. 3b+4b+5b = 3*4 12b = 12 b = 1 unit. b is the radius of the inscribed circle. tanc = 3/4 c = atan(¾)° d = ½(180-atan(¾)) d = 71.5650511771° tane = 4/3 e = atan(4/3)° f = ½(180-atan(4/3)) f = 63.4349488229° sin71.5650511771 = 1/g g = 1.0540925534 units. sin63.4349488229 = 1/h h = 1.1180339888 units. j = 90-71.5650511771 j = 18.4349488229° k = 90-63.4349488229 k = 26.5650511771° l = 180-26.5650511771-18.4349488229 l =...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2514912

Let the bigger square side length be 1 unit. a² = 2(1)² a = √(2) units. a is the diagonal of a bigger square. Let b be the side length of the big square. c = (1+b) units. It implies; Calculating b. √(2)² = (1+b)²+b² 2 = 1+2b+b²+b² 2b²+2b-1 = 0 Resolving the quadratic equation via completing the square approach to get b, side length of the big square. 2b²+2b-1 = 0 b²+b = ½ (b+½)² = ½+(½)² (b+½)² = ¼(3) b = -½±½√(3) It implies; b = ½√(3)-½ b = ½(√(3)-1) units...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2513313

Let the side length of the regular pentagon be 1 unit. a² = 1+0.5²-2*0.5cos108 a = 1.2486060205 units. (1.2486060205/sin108) = (0.5/sinb) b = 22.3861775592° c = 180-108-b c = 49.6138224408° d = c-b d = 49.6138224408-22.3861775592 d = 27.2276448816° e = 180-2c e = 180-2*49.6138224408 e = 80.7723551184° f = 180-27.2276448816-e f = 72° (1.2486060205/sin72) = (g/sin27.2276448816) g = 0.6006698572 units. h² = 0.6006698572²+0.5²-2*0.6006698572*0.5cos49.6138224408...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2513053

tana = 10/5 a = atan(2)° b = 180-2a b = (180-2atan(2))° c² = 5²+5²-2*5*5cos(180-2atan(2)) c = 4.472135955 units. d = ½(90-(180-2atan(2))) d = atan(2)° e = 90-d e = 26.5650511771° Shaded Area is; ½*4.472135955*10sin26.5650511771 = 22.146600355 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2513010

Let the side length of the regular pentagon be 1 unit. a = 108-90 a = 18° 108° is the single interior angle of the regular pentagon. b = 90-18 b = 72° c =108-72 c = 36° sin18 = d/1 d = 0.3090169944 units. e² = 1+0.3090169944²-2*0.3090169944cos36 e = 0.7716809592 units. (0.7716809592/sin36) = (0.3090169944/sinf) f = 13.6138224422° g² = 1²+0.5²-2*0.5cos108 g = 1.2486060205 units. (1.2486060205/sin108) = (0.5/sinh) h = 22.3861775592° j = 108-13.6138224422-h...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2512953

a = ⅐*180(7-2) a = ⅐(900)° a is the single interior angle of the regular heptagon. (√(2)/sin(900/7)) = (1/sinb) b = 33.5619539118° c = 180-(900/7)-b c = 17.8666175168° (d/sin17.8666175168) = (√(2)/sin(900/7)) d = 0.5549581321 units. e = 1-d e = 0.4450418679 units. f² = 2-2cos(900/7) f = 1.8019377358 units. g = ½(180-(900/7)) g = 25.7142857143° h = (900/7)-g h = 102.8571428571° Therefore, length AB is; Let it be j. j² = 1.8019377358²+0.4450418679²-2*1.80193...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2512264

tan60 = a/2 a = 2√(3) cm. b² = 2²+(2√(3))² b² = 16 b = 4 cm. Let the side length of the ascribed regular triangle (equilateral triangle) be x. c = (x-2) cm. d = ½(x-2) cm. e = (x-4) cm. f = x-(x-4)-½(x-2) f = ½(2x-2x+8-x+2) f = ½(10-x) cm. sin60 = g/½(10-x) g = ¼(10-x)√(3) cm. cos60 = h/½(10-x) h = ¼(10-x) cm. j = 2√(3)-¼(10-x)√(3) j = ¼(8√(3)-10√(3)+√(3)x) j = ¼(√(3)x-2√(3)) cm. sin60 = k/(x-2) k = ½(√(3)x-2√(3)) cm. It implies; ½*½(√(3)x-2√(3))*¼(...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2511387

Let BC be 1 unit. a = 180-54-21-15-33 a = 57° (1/sin57) = (b/sin(54+21)) b = 1.1517344989 units. c = 180-21-33 c = 126° (1/sin126) = (d/sin21) d = 0.4429671466 units. f² = 0.4429671466²+1.1517344989²-2*0.4429671466*1.1517344989cos15 f = 0.7328841114 units. Therefore, the required angle, x is; (0.7328841114/sin15) = (0.4429671466/sinx) x = 9°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2511307

Let the side length of the square be 2a cm. b² = a² b² = ½(a)² b = ½√(2)a cm. b is the radius of the circle. c² = 2(2a)² c² = 8a² c = 2√(2)a cm. c is the diagonal of the square. d = c-b d = 2√(2)a-½√(2)a d = ½(3√(2)a) cm. e = PA+PB e = 5+4 e = 9 cm. Therefore; 9² = (½√(2)a)²+(½(3√(2)a)²-2*(½√(2)a)(½(3√(2)a)cos135 81 = 0.5a²+4.5a²-3a²cos135 81 = 5a²+2.1213203436a² 81 = 7.1213203436a² a = √(11.3742952279) a = 3.3725799068 cm. 2a (side length of the square) is;...

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