Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3024978

Let a be the side of the square. cos30 = b/8 b = 4√(3) cm. sin30 = c/8 c = 4 cm. a = b+12 a = (12+4√(3)) cm. Area Square is; a² = (12+4√(3))² = 358.27687752661 cm² d² = 12²+4²-2*4*12cos30 d = 8.76707255797 cm. (8.76707255797/sin30) = (12/sine) e = 180-43.18678543201 e = 136.81321456799° f = 360-e-180 f = 43.18678543201° g² = 8²+8.76707255797²+2*8*8.76707255797cos43.18678543201 g = 6.21165708246 cm. (6.21165708246/sin43.18678543201) = (8.76707255797/sinh)...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3023335

Let the single side length of the two congruent inscribed regular pentagon be 1 unit. Area Green is; 0.5*5(1/(2tan(180/5))) = 1.72047740059 square units. Calculating Area Shaded. a = ⅕(180*3) = 108° Where a is the single interior angle of the congruent inscribed regular pentagon. b² = 2-2cos108 b = 1.61803398875 units. c = 1+b c = 2.61803398875 units. d² = 2.61803398875²+1²-2*1*2.61803398875cos72 d = 2.49721204096 units. Where d is the radius of the inscribed...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3021112

Sir Mike Ambrose is the author of the question. Area Orange exactly in square cm decimal is; Area triangle with height 1.64419338427 cm and base (4.95325421888sin129.5411167921) cm = ½*1.64419338427*(4.95325421888sin129.5411167921) = 3.14023734337 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3019244

a = atan(3/2)° b = atan(2/3)° b = Angle BEG. tan(atan(2/3)) = c/2 ⅔ = c/2 c = (4/3) cm. c = BG. OG = 2-(4/3) OG = ⅔ cm. Notice; Radius of the circle is 2 cm. Angle OGH = 180-atan(3/2) = 123.69006752598° (2/sin123.69006752598) = ((⅔)/sind) d = 16.10211375199° d = Angle OHG. Angle GOH = 180-16.1021137519-123.69006752598 = 40.20781872212° It implies; Length GH is; (GH/sin40.20781872212) = ((2/3)/sin16.10211375199) GH = 1.55173771485 cm.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3017238

Sir Mike Ambrose is the author of the question Let the side of the regular Pentagon be 2 units. Therefore; Area orange is; Area triangle with height 1.7013016167 units and base 1.8345763534sin(18) units = ½*1.7013016167*1.8345763534sin(18) = 0.48224693327 square units. Area blue is; Area trapezoid with parallel sides 0.21654236465 units and 1 unit, and height 1.9021130326 units. ½*1.9021130326(0.21654236465+1) = 1.15700054326 square units. It implies; Area Orang...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3015837

a² = 2*12²-2*12²cos120 a = 12√(3) cm. a = 20.78460969083 cm. a = AB. b = 2a b = 24√(3) cm. b = 41.56921938165 cm. b = AC. c² = 12²+41.56921938165²-24*41.56921938165cos30 c = 31.74901573277 cm. c = CO1 (31.74901573277/sin30) = (12/sind) d = 10.89339464913° d = Angle O1CA. e = 180-10.89339464913-30 e = 139.10660535087° e = Angle AO1C. f = asin(12/31.74901573277) f = 22.2076542986° f = Angle O1CD. cos22.2076542986 = g/31.74901573277 g = 29.39387691339 cm. g = C...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3014345

a² =2*12² a = 12√(2) cm. a = AC. 12b+12b+12√(2)b = 12*12 b = 144/(24+12√(2)) b = 12-6√(2) cm. b = 3.51471862576 cm. b = radius of the inscribed circle. BF = ½(12√(2)) BF = 6√(2) cm. c² = (6√(2))²+3.51471862576²-12√(2)*3.51471862576cos45 c = 6.49435320175 cm. c = EF. (6.49435320175/sin45) = (6√(2)/sind) d = 112.5° d = Angle BEF. e = 2(180-112.5-45) e = 45° f = 112.5-90 f = 22.5° cos22.5 = 12/g g = 12.98870640351 cm. g = EG. FG = 12.98870640351-6.494353201...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3012696

a² = 15²+12²-2*12*15cos120 a = 23.43074902772 cm. Where a is the side length of the equilateral triangle ABC. (23.43074902772/sin120) = (12/sinb) b = 26.32950349168° c = 60-b c = 33.67049650832° d² = 15²+15²-2*15*15cos120 d = 25.98076211353 cm. e = atan(7.5/25.98076211353) e = 16.10211375199° f = 90-e f = 73.89788624801° g = 180-f-c g = 72.43161724367° (15/sin72.43161724367) = (h/sin73.89788624801) h = 15.11661227595 cm. i = a-h i = 8.31413675177 cm. Let...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3011353

Sir Mike Ambrose is the author of the question. Area Green exactly is; Area trapezoid with parallel sides 3 cm and (6-2√(3)) cm, and height (6-3√(3)) cm. = ½(6-3√(3))(3+(6-2√(3))) = ½(6-3√(3))(9-2√(3)) = ½(72-39√(3)) cm² = 3(24-13√(3))/2 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3009704

a² = 12²-6² a = 6√(3) cm. a = 10.39230484541 cm. a = circle's radius = AD = AG. b² = 2*10.39230484541²-2*10.39230484541²cos120 b = 18 cm. b = DG. c = 12-10.39230484541 c = 1.60769515459 cm. c = CD. d² = 12²+1.60769515459²-24*1.60769515459cos60 d = 11.28239078631 cm. d = BD. (11.28239078631/sin60) = (1.60769515459/sine) e = 7.088663081° e = Angle CBD. f = 60-7.088663081 f = 52.911336919° f = Angle DBG. (12/sing) = (10.39230484541/sin52.911336919) g = 112.9113369...

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