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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2508440

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Considering the quadrilateral cyclic. a² = 2*10²-2*10²cosx a² = 200-200cosx --- (1). a² = 9²+7²-2*9*7cosx a² = 130+126cosx --- (2). Equating (1) and (2). a² = a² 200-200cosx = 130+126cosx 200-130 = 126cosx+200cosx 326cosx = 70 x = acos(70/326) x = 77.6006688181° y = 180-x y = 102.3993311819° Therefore, area cyclic quadrilateral is; Area triangle with height 10 units and base 10sin77.6006688181 units + Area triangle with height 9 units and base 7sin102.3993311819...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2506798

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Let the base of the ascribed triangle be 1 unit. a = 180-12-42-12-36 a = 180-102 a = 78° (1/sin(12+36)) = (b/sin78) b = 1.3162274259 units. c = 180-12-42 c = 180-54 c = 126° (1.3162274259/sin126) = (d/sin12) d = 0.3382612127 units. f² = 0.3382612127²+1-2*0.3382612127cos12 f = 0.6728163648 units. It implies, the required angle x, is; (0.6728163648/sin12) = (0.3382612127/sinx) x = 6.0442188698°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2506755

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Let AC be 2 units. a = 180-15-30 a = 135° b = 180-135 b = 45° (1/sin15) = (c/sin30) c = 1.9318516526 units. d² = 1.9318516526²+1²-2*1.9318516526*1cos45 d = √(2) units. d = 1.4142135624 units. Therefore, the required angle x is; (1.4142135624/sin45) = (1/sinx) x = 30°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2506708

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a² = 2²+1² a = √(5) units. b² = 2(a)² b² = 2√(5)² b² = 10 b = √(10) units. b is the diagonal of the blue square. tanc = (2/1) c = atan(2)° d = c-45 d = (atan(2)-45)° Calculating x. It implies; x² = 1+√(10)²-2√(10)*1cos(atan(2)-45) x = 2.2360679775 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2506656

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a = ½(PQ) a = ½(4) a = 2 units. b = 90-60 b = 30° c = 180-45-30 c = 105° d = 180-105-60 d = 15° It implies; (2/sin105) = (e/sin15) e = 0.5358983849 units. Therefore, Red Painted Area is; 4(area triangle with height 2 units and base 0.5358983849sin60 units) = 4(0.5*2*0.5358983849sin60) = 1.8564064607 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2506253

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Calculating the area of the inscribed square. Let the inscribed square length be a. It implies; b = 3-2 b = 1 unit. Therefore; The Area of the inscribed square (a²) is; a² = 2²+1² a² = 5 square units.

We are smart and clever!

https://onlineedumath.com/update/we-are-smart-and-clever/2505977

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Our amazing learners back in the day, 2013 From left to right; Alamin, Esther, Favour, Victoria, Mufida and Amina. We are proud of you all.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2505483

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Calculating R, radius of the four inscribed equal half circles. Notice! Radius of the four equal inscribed circles is 1 unit. Therefore; (2R-1)² = 1²+a² 4R²-4R+1 = 1+a² a = √(4R²-4R) units. It implies; (R+1)² = a²+(R-1)² (R+1)² = √(4R²-4R)²+(R-1)² R²+2R+1 = 4R²-4R+R²-2R+1 8R = 4R² 4R = 8 R = 2 units. 2R = 2*2 = 4 units. Where 2R = 4 units is the radius of the ascribed circle. Therefore, area yellow is; Area circle with radius 4 units - 2(area circle with radius 2...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2505441

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55+2(90)+270+a = 180(5-2) a+505 = 540 540° is the sum of the interior angles of a pentagon. It implies; a = 540-505 a = 35° tan55 = b/4 b = 5.712592027 cm. c = b-4 c = 1.712592027 cm. tan55 = 8/d d = 5.6016603057 cm. e = d-4 e = 1.6016603057 cm. It implies, area ABCDE is; Area triangle with height 4 cm and base 5.712592027 cm + Area trapezoid with parallel sides 4 cm and 1.712592027 cm respectively and height 1.6016603057 cm. = ½*4*5.712592027+½*(4+1.712592027)*1...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2504327

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a² = 4²+6²-2*4*6cos60 a = 5.2915026221 units. (5.2915026221/sin60) = (4/sinb) b = 40.8933946491° c² = 6²+2²-2*6*2cos60 c = 5.2915026221 units. (5.2915026221/sin60) = (2/sind) d = 19.1066053509° Therefore, the required angle, x is; x = b+d+60 x = 40.8933946491+19.1066053509+60 x = 120°
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