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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2966130

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a² = 4²+4²-2*4*4cos150 a = 7.72740661031 cm. Where a is BE. b = angle ABE = ½(180-150) b = 15° c = angle AFE = 180-45-15 c = 120° (4/sin120) = (d/sin45) d = 3.26598632371 cm. Where d is EF. Notice; EF = CF Therefore; Length CF = 3.26598632371 cm.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2966102

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tan15 = a/4 a = 1.07179676972 cm. Where a is AD. b² = 1.07179676972²+4² b = 4.14110472164 cm. Where b is DF. c = 4-1.07179676972 c = 2.92820323028 cm. Where c is CD. sin60 = d/2.92820323028  d = 2.53589838487 cm. Where d is DE = BE. e² = 16+16-2*16cos150 e = 7.72740661031 cm. Where e is BF. f² = 7.72740661031²+2.53589838487²-2*2.53589838487*7.72740661031cos45 f = 6.19925645689 cm. Where f is EF. (6.19925645689/sin45) = (2.53589838487/sing) g = 16.81321456806° Where g is angle BFE. h = 1...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2964663

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a = acos(3/5) Where a is angle GCA. tan(acos(3/5)) = b/3 b = 4 units. Where b is AG. c = atan(⅔) Where c is angle GBE. d = 180-acos(3/5)-atan(⅔) d = 93.17983011986° Where d is angle BEC. (6/sin93.17983011986 ) = (e/sin(acos(⅗))) e = 4.80740170062 units. Where e is BE. Area Parallelogram BCDE is; 2(area triangle with height 4.80740170062 units and base 6sin(atan(⅔)) units. = 2(0.5*4.80740170062*6sin(atan(⅔))) = 16 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2964658

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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/2962920

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a = 15-11.25  a = 3.75 m. b = atan(3/3.75)° c = 90-atan(3/3.75) c = 51.34019174591°  Calculating Length CD. tan(51.34019174591) = 15/CD CD = 15/tan(51.34019174591)  CD = 12 m.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2962888

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Let the square side be a. tan60 = a/b b = a/tan60 cm. Where b is BS. Calculating a. 2b+a = 6 2(a/tan60)+a = 6 a((2/tan60)+1) = 6 a = 6/((2/tan60)+1) a = 6(2√(3)-3) cm. Area Shaded STR is; ½(a²) = ½(6(2√(3)-3))² = 54(7-4√(3)) cm² = 3.87702556513 cm²

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.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2959686

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Let the small inscribed square side be a. Let b be the side of the big inscribed square. It implies; b = 10-a Therefore; Calculating a. tan(90-75) = a/(10-a) a(1+tan15) = 10tan15 a = (10tan15)/(1+tan15) a = ⅓(15-5√(3)) cm. b = 10-⅓(15-5√(3)) b = ⅓(15+5√(3)) cm. Blue area is; a²+b² = (⅓(15-5√(3)))²+(⅓(15+5√(3)))² = ⅓(200) cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2958211

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Sir Mike Ambrose is the author of the question. Area R exactly in its simplest decimal form is; Area triangle with height 4.54406194214 units and base 2sin(atan(2)) units. = ½*4.54406194214*2sin(atan(2)) = 4.54406194214*sin(atan(2)) = 4.06433255864 square units.
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