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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1886518

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tan(0.5atan(3/4))= a/24 a = 8 units. Where a is DF. b = 18-8 b = 10 units.  Where b is CF. tan(0.5atan(4/3))= c/18 c = 9 units. Where c is BE. d = 24-9 d = 15 units. Where d is CE. It implies; (EF)² = (CF)²+(CE)² (EF)² = 10²+15² EF = √(325) EF = 5√(13) units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1881048

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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/1879295

mathematics-question-and-solution_16864104766pxmz2.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. 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° Where g is angle PSA. ta...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1877260

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AD = √(12²+6²) AD = 6√(5) cm. Angle AMD = atan(2√(5))° Angle EBM = (90-ata(2√(5)))° sin(90-ata(2√(5))) = EM/9 EM = 1.96396101212 cm. Angle CME = 180-atan(2√(5)) = 102.60438264838° (CE)² = 9+1.96396101212²-6*1.96396101212cos102.60438264838 CE = 3.92792202425 cm. (3.92792202425/sin102.60438264838) = (1.96396101212/sina) a = 29.20593224736° Where a is angle ECM. cos29.20593224736 = 6/CF CF = 6.87386354243 cm. EF = CF-CE  EF = 6.87386354243-3.92792202425 EF = 2.94594151818 cm. Therefore;...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1854583

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BC = √(3²+4²) BC = 5 cm. tana = (4/3) a = atan(4/3)° tan(½(atan(4/3))) = AF/3 AF = 3tan(½(atan(4/3))) AF = 1.5 cm. b = 90-½(atan(4/3)) b = 63.43494882292° Where b is angle AFE. sin63.43494882292 = AE/1.5 AE = 1.3416407865 cm. (BE)² = 3²-1.3416407865² BE = 2.683281573 cm. Calculating Length CE. (CE)² = 5²+2.683281573²-2*5*2.683281573cos(0.5atan(4/3))  CE = 2.86356421266 cm. (2.683281573/sinc) = (2.86356421266/sin(0.5atan(4/3))) c = 24.77514056879° Where c is angle BCE. Angle ECF = atan...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1847527

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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.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1842275

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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/1810280

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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. Let the radius of the green ins...

Mathematics Question and Solution

https://onlineedumath.com.free/update/mathematics-question-and-solution/1807684

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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 = 20.78460969082 units. l² = (2√(129)...

Mathematics Question and Solution

https://onlineedumath.com.free/update/mathematics-question-and-solution/1797707

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Calculating r, radius of the circle. (r/tan40)+(r/tan30) = 10 r = 3.42020143326 cm. (a/sin60) = (10/sin40) a = 13.47296355334 cm. tan40 = 3.42020143326/b b = 4.07603734548  (4.07603734548/sin50) = (c/sin80) c = 5.24005260459 cm. d = 13.47296355334-4.07603734548 d = 9.39692620786 cm. e² = 9.39692620786²+5.24005260459²-2*9.39692620786*5.24005260459cos130 e = 13.38142726359 cm. (13.38142726359/sin130) = (5.24005260459/sinf) f = 17.45618925878° g = 100-17.45618925878  g = 82.54381074122°...
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