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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3058839

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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. sin67.58908946897 = 19.2148198619/k k = 20....

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3051425

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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. L...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3048366

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a² = 12²+6² a = 6√(5) cm. a = AE. b = atan(1/2)° b = angle DAE. c = 180-atan(2) c = 116.56505117708° c = angle AFC. d = 180-116.56505117708-atan(1/2) d = 36.86989764584° (6/sin36.86989764584) = (e/sin116.56505117708) e = 8.94427191 cm. f = 6√(5)-8.94427191  f = 4.472135955 cm. g = 180-2atan(2) g = 53.13010235416° g = angle AEB. cos53.13010235416 = h/4.472135955 h = 2.683281573 cm. It implies; Area S1 is; 0.5*2.683281573*4.472135955sin53.13010235416...

Mathematics Questions and Solution

https://onlineedumath.com/update/mathematics-questions-and-solution/3032283

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Sir Mike Ambrose is the author of the question. Let the side of the inscribed square be 2 units. Therefore; Area S is; Area triangle with two side 1.03528 units and 0.53589838486 units, and angle 45° = 0.5*1.03528*0.53589838486sin45 = 0.1961531464 space units. Area R is; Area triangle with two side (2√(2)-2) units and 0.65961372961 units, and angle 72.367664609° = 0.5*(2√(2)-2)*0.65961372961sin72.36766460 = 0.26038499708 space units. Area T is; Area sector with radius 2√(2) units a...

Mathematics Question and Solution

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

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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/3012696

mathematics-question-and-solution_1741392176954Hkq.jpeg
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/2977546

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Let the inscribed square side be a. tan60 = a/b a = √(3)b cm. Where b is BS = CR It implies; b+√(3)b+b = 10 (2+√(3))b = 10 b = 10(2-√(3)) cm. Therefore; a = √(3)*10(2-√(3)) a = 10(2√(3)-3) cm. a = 4.64101615138 cm. sin60 = 10(2√(3)-3)/c c = 20(2-√(3)) cm. Where c is CQ. d² = 10²+(10(2-√(3)))²-20*10(2-√(3))cos60 d = 8.96575472168 cm. Where d is AS. (8.96575472168/sin60) = (10(2-√(3))/sine) e = 15° Where e is angle PAS. f = 180-60-15 f = 105° Where f is angle BSA. g = 105-90 g = 15° Whe...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2974204

mathematics-question-and-solution_1739323824vaVWsU.jpeg
Let the inscribed square side be a. tan60 = a/b a = √(3)b cm. Where b is BS = CR It implies; b+√(3)b+b = 10 (2+√(3))b = 10 b = 10(2-√(3)) cm. Therefore; a = √(3)*10(2-√(3)) a = 10(2√(3)-3) cm. a = 4.64101615138 cm. sin60 = 10(2√(3)-3)/c c = 20(2-√(3)) cm. Where c is CQ. d² = 10²+(10(2-√(3)))²-20*10(2-√(3))cos60 d = 8.96575472168 cm. Where d is AS. (8.96575472168/sin60) = (10(2-√(3))/sine) e = 15° Where e is angle PAS. f = 180-60-15 f = 105° Where f is angle BSA. g = 105-90 g = 15° Whe...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2961311

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a² = 10²+10²-2*10*10cos150 a = 19.31851652578 cm. Where a is AF which is equal AE. (19.31851652578/sin150) = (10/sinb) b = 15° Where b is angle BAF. tan15 = c/10 c = 2.67949192431 cm. It implies; Area Green Region is; Area square with side 10 cm - 2(area triangle with height 2.67949192431 cm and base 10 cm. = 10² - ½*2(10*2.67949192431) = 100-26.7949192431 = 73.2050807569 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2961298

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a = atan(2)° b = 180-2a b = (180-2atan(2))° Where b is angle FBP. Notice; CQ = 5 cm. Calculating FP using sine rule. Let FP be c. (c/sin(180-2atan(2))) = (5/sin(atan(2))) c = 4.472135955 cm. Therefore; PQ is; CF-CQ-FP = √(10²+5²)-5-4.472135955  = (3√(5)-5) cm. = 1.7082039325 cm.
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