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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1911304

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a = 180-2*8 a = 164° Where a is angle AOT, O is the centre of the circle. Angle ABT is; ½(360-164) = ½(196) = 98° Angle BAT is; ½(180-98) = 41° Therefore; Angle ACT is; 180-41-(180-98) = 180-41-82 = 180-123 = 57°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1910716

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a = asin(1/6) b = 2a b = 2(asin(1/6)) Where b is angle BAC. sin(2(asin(1/6))) = c/6 c = ⅓(√(35)) units. c = 1.97202659437 units. Where c is BE. d = 90-2asin(1/6) d = 70.81186354628° Where d is angle ABE. cos70.81186354628 = e/1.97202659437 e = 0.64814814815 unit. Where e is BD. tan70.81186354628 = f/0.64814814815 f = 1.86246956135 units. Where f is DE. tan(acos(1/6)) = g/0.64814814815 g = 3.83449615572 units. Where g is DF. h = 3.83449615572-1.86246956135 h = 1.97202659437 units. Where...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1910693

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Let the side length of the regular heptagon be 1 unit. a = ⅐(180*5) a = ⅐(900)° b = 0.5(360-2(900/7)) b = ⅐(360)° c = 2cos(360/7)+1 c = 2.24697960372 units. d² = 2-2cos(90/7) d = 1.8019377358 units. Where d is the radius of the arc. Area Orange is; 0.5*2.24697960372*1sin(360/7)+0.5*sin(900/7) = 1.26929543856 square units. Let e be alpha. Calculating e. e = 0.5((900/7)-(360/7)) e = ⅐(270)° (1.8019377358/sin(270/7)) = (2.24697960372/sinf) f = 128.96966294...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1908221

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a² = 11*11+10*10-2*11*10cos50 a = 8.92113926968 cm. (8.92113926968/sin50) = (10/sinb) b = 59.16920318624° c = 180-50-59.16920318624 c = 70.83079681376°  d = 0.5a d = 4.46056963484 cm. e² = 4.46056963484²+5.5²-2*4.46056963484*5.5cos59.16920318624 e = 5 cm. (5/sin59.16920318624) = (5.5/sinf) f = 70.83079681376° g² = 4.46056963484²+5²-2*4.46056963484*5cos70.83079681376 g = 5.5 cm. (5.5/sin70.83079681376) = (5/sinh) h = 59.16920318624° i = 180-59.16920318624-70.83079681376 i = 50° (j/tan...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1908220

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Let the side length of the regular pentagon be 1 unit. tan36 = a/0.5 a = 0.363271264 unit. b² = 0.363271264²+0.5² b = 0.61803398875 unit. tan72 = c/0.5 c = 1.53884176859 units. Area Green is; 0.5sin108+0.5*1.53884176859*0.5-0.5*0.5*0.363271264-0.5*0.61803398875*0.61803398875sin36 = 0.65716389015 square units. Calculating Area Blue. Let d be the radius of the inscribed blue circle. Calculating d. e = 0.5(90+36) e =63° (d/tan63) + (d/tan45) = a d = 0.363271264/((1/tan63)+(1/tan45)) d...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1908205

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a = asin(1/6) b = 2a b = 2asin(1/6) Where b is angle BAC cos(2asin(1/6)) = c/6 c = ⅓(17) units. Where c is AD. d = 6-c d = 6-⅓(17) d = ⅓ units Where d is CD. sin(2asin(1/6)) = e/6 e = 1.97202659437 units. Where e is BD. f = asin((1/3)/2) f = 9.59406822686° Where f is angle EBC = angle CAE. tan(9.59406822686) = g/(17/3) g = 0.95784148869 units. Or  g = 17√(35)/105 units. Where g is DE.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1907223

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Let the radius of the ascribed circle be 2 units. sin10 = a/2 a = 0.34729635533 units. a is O1D. cos10 = b/2 b = 1.96961550602 units. b is CD. Let the small inscribed circle radius be r. Calculating r. (2-r)² = r²+(0.34729635533+r)² 4-4r+r² = r²+0.12061475843+0.69459271066r+r² r²+4.69459271066r-3.87938524157 = 0 It implies; r = 0.716881 units. c = 2-0.34729635533 c = 1.65270364467 units. c is BD. tand = 1.65270364467/1.96961550602 d = 40° d is angle BCD. tane = 0.716881/1.96961550602...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1906109

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Let the small square side length be 1 unit. (a/sin30) = (1/sin15) a = 1.93185165258 units. a is the big square side length. b² = 1.93185165258²+1-2*1.93185165258cos45 b = √(2) units. b = 1.41421356237 units. (1.93185165258/sinc) = (√(2)/sin45) c = 75° Therefore required angle is; 180-c = 180-75 = 105°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1904531

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Radius of the circle r, is; πr² = 25π r = 5 cm. Therefore diameter is; 2r = 10 cm. Let the side of the small yellow Inscribed square be x. a² = 2x² a = √(2)x cm. Where a is a diagonal of the small yellow Inscribed square. Extending a. We now have a rectangle inscribed the circle (cyclic quadrilateral) with diameter 10 cm. Where 6 cm is the width and (6+√(2)x) cm is the length. Calculating x (small yellow Inscribed square length) 10² = 6²+(6+√(2)x)² 100-36 = (6+√(2)x)² 8 = 6+√(2)x 2 = √...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1904180

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r = √(3²+4²) r = 5 units. r is the radius of the quarter circle. (5-x)² = x²+(3+x)² x is the radius of the inscribed yellow circle. Therefore; 25-10x+x² = x²+9+6x+x² x²+16x-16 = 0 (x+8)² = 16+64 x = -8±√(80) x = -8±4√(5) Therefore; x ≠ -8-4√(5) x = -8+4√(5) x = 4√(5)-8 units. x = 0.94427191 units. Area yellow Inscribed circle is; πx² = π(4√(5)-8)² = 80-64√(5)+64 = (144-64√(5))π square units. = 16π(9-4√(5)) square units. = 0.89164944001π square units.
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