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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1936683

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Let CE be 2 units. Let the square side be a. b =√(2)a unit. c = (√(2)a-2) units. Calculating a  2²+2² = (√(2)a-2)² 8 = 2a²-4√(2)a+4 2a²-4√(2)a-4 = 0 a²-2√(2)a-2 = 0 (a-√(2))² = 2+2 a = √(2)±2 Therefore; a ≠ (√(2)-2) units. a = (√(2)+2) units. a is AD. tand = (2/(√(2)+2)) d = atan(2/(√(2)+2))° d = 30.36119340482° d is angle ABF. (BG)² = 1+3.41421356237²-2*1*3.41421356237cos45 BG = 2.79793265193 units. (2.79793265193/sin45) = (1/sine) e = 14.63880659519° e is angle CBG. Calculating re...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1935913

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a² = (2r)²+r² a = √(5r²) a = √(5)r units. b = a-r b = (√(5)r-r) units. Considering similar triangle ratios. Calculating r.  (√(5)r-r) = 2r √(2) = r Cross Multiply.  (√(5)r-r) = 2√(2) r(√(5)-1) = 2√(2) r = (2√(2))/(√(5)-1) r = ½(√(2)+√(10)) units. r = 2.28824561127 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1935892

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Let the base of the composite triangle be 1 unit. a = 180-3-6 a = 171° (1/sin171) = (b/sin3) b = 0.33455515209 units. c = 180-30-87 c = 180-117 c = 63° (d/sin30) = (1/sin63) d = 0.56116311882 units. e² = 0.56116311882²+0.33455515209²-2*0.56116311880.334555152092*0.33455515209cos81 e = 0.60670683106 units. Calculating x. (0.60670683106/sin81) = (0.33455515209/sinx) x = 33°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1935400

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i = k  And i = 180-101-39 = 40° Therefore; k = 40° While j = 101° Or Calculating i. i = 90-11-39 i = 90+50 i = 40° a = 180-2(11) a = 180-22 a = 158° b = 360-a b = 360-158 b = 202° Calculating j. j = ½(b) j = ½(202) j = 101° Calculating k. k+j+39 = 180 k+101+39 = 180 k = 180-140 k = 40°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1935378

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Considering similar triangle ratios. 3 = (3+x) y = (4+y) Where x and y are the width and length of the inscribed rectangle ascribing the red region respectively. Cross Multiply. 12+3y = 3y+xy xy = 12 square units. Therefore; Area Red Region is; ½(xy) = ½(12) = 6 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1935205

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a = ½(180-150) a = 15° b = 90-15 b = 75° (c/sin45) = (12/sin60) c = 9.79795897113 cm. (12/sin60) = (d/sin45) d = 9.79795897113 cm. e² = 9.79795897113²+12²-24*9.79795897113cos135  e = 20.15631110909 cm. (12/sinf) = (20.15631110909/sin135) f = 24.89609063899° g = 120-24.89609063899 g = 95.10390936101° h = 180-24.89609063899-75 h = 80.10390936101° (9.79795897113/sin80.10390936101) = (i/sin75) i = 9.60704430088 cm. Therefore; Area Shaded Region is; 0.5*9.60704430088*9.79795897113sin95....

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1935082

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Let the side of the regular pentagon be 1 unit. a = 108-90 a = 18° b = 180-108-18 b = 54° (1/sin54) = (c/sin108) c = 1.17557050458 units. (1/sin54) = (d/sin18) d = 0.38196601125 unit. e = 1-d e = 0.61803398875 unit. f² = 0.61803398875²+1²-2*0.61803398875cos108 f = 1.3281310261 units. (1.3281310261/sin108) = (1/sing) g = 45.73230144788° h = 180-108-45.73230144788 h = 26.26769855212°  I = 180-72-45.73230144788 i = 62.26769855212° (0.61803398875/sin62.2676...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1934249

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Let BC be 1 unit. a = 180-30-18 a = 132° a is angle BDC. (1/sin132) = (b/ sin18) b = 0.41582338164 units. b is BD. c = 180-72 c = 108° c is angle BAC. (1/sin108) = (d/sin36) d = 0.61803398875 units. d is AB. (AD)² = 0.61803398875²+0.41582338164²-2*0.61803398875*0.41582338164cos6 AD = 0.20905692653 units. Calculating x. (0.20905692653/sin6) = (0.41582338164/sinx) x = 12°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1933349

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(m+n)² = c²+b²-2bccosA --- (1). h² = c²-m²  c² = h²+m² --- (2). h² = b²-n²  b² = h²+n² --- (3). Substituting (2) and (3) in (1). (m+n)² = h²+m²+h²+n²-2bccosA m²+2mn+n² = 2h²+m²+n²-2bccosA 2mn = 2h²-2bccosA 2h² = 2mn-2bccosA Therefore; h² = mn+bccosA Proven!

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1933272

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Let radius of the half circle be 2 units. sin22.5 = d/2 d = 0.76536686473 units. BZ = 2d BZ = 1.53073372946 units. e = 90-½(180-45) e = 22.5° cos22.5 = f/4 f = 3.69551813005 units. f is CZ. cos67.5 = g/1.53073372946 g = 0.58578643763 units. h = 2g h = 1.17157287525 units. j = 0.5(2-1.17157287525) j = 0.41421356238 units. CX = 2+0.41421356238  CX = 2.41421356238 units. Calculating c. tan22.5 = c/2.41421356238 c = 1 unit. Calculating b. 2² = k²+0.4...
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