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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1919356

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Notice; Triangle ABC is equilateral. Let the side of the regular pentagon be a. Single interior angle of the regular pentagon is; ⅕(180*3) = 108° (HF)² = 2a²-2a²cos108 HF = 1.61803398875a units. sin18 = b/a b = 0.30901699437a units. c = HF-b c = 1.61803398875a-0.30901699437a c = 1.30901699438a units. d = 1-c d = (1-1.30901699438a) units. cos18 =e/a e = 0.9510565163a units. Calculating a, side length of the regular pentagon. tan60 = e/d √(3) = 0.9510565163a/(1-1.30901699438a) √(3)-2.26...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1919162

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Let AB be 1 unit. (1/sin40) = (BC/sin70) BC = 1.46190220008 units. (1.46190220008/sin120) = (AE/sin40) AE = 1.08506357513 units. (1.08506357513/sin40) = (CE/sin20) CE = 0.57735026919 units. BE = 1.46190220008-0.57735026919 BE = 0.88455193089 units. (BD)² = 0.88455193089²+0.57735026919²-2*0.57735026919*0.88455193089cos60 BD = 0.77786191343 units. Therefore angle alpha (angle BDE) is; Let it be x. (0.88455193089/sinx) = (0.77786191343/sin60) x = 80° Angle alpha (required angle) = 80°

We are smart and clever!

https://onlineedumath.com/update/we-are-smart-and-clever/1918927

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Our certified, experienced, resilient, productive and committed Math Educators will be glad to be at your service. We make teaching and learning Mathematics fun for learners, we expose the simplicity and beauty of Mathematics to our learners, and mentor them to appreciate and love Mathematics. Communicate us to mentor/educate you Mathematics better online at affordable tuition. We are smart and clever.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1918922

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a² = 12²-6² a = 6√(3) cm. Where a is BD. b = ½(a) b = 3√(3) cm. Where b is BG = DG. c = atan(6/(3√(3)) c = 49.10660535087° Where c is angle AGD = angle BGE. d = 60-49.10660535087 d = 10.89339464913° Where d is angle BAE. e = 180-60-10.89339464913 e = 109.10660535087° Where e is angle BEA. (12/sin109.10660535087) = (f/sin60) f = 10.99818166789 cm. Where f is AE. (12/sin109.10660535087) = (g/sin10.89339464913) g = 2.4 cm. Where g is BE. h = 12-2.4 h = 9.6 cm. Where h is CE. i² = (6√(3))²...

We are smart and clever!

https://onlineedumath.com/update/we-are-smart-and-clever/1918910

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Screenshot of yesterday's teaching and learning activities with smart and clever Oghenevwarhe (year 8), Oghenemano (year 8), Ogheneruno (year 9), Destiny (year 10) and Ogheneruona (year 10). Educators: Mr. Ogheneovo, Mr. Mudiaga and Mr. Daniel We are smart and clever.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1916952

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Let AB be 5 units. Radius, r of the inscribed circle is; 5*(2/5) r = 2 units. a = ⅛(180*6) a = 135° 2b² = 25 b = ½(5√(2)) units. c = 180-0.5(135)-45 c = 67.5° sin22.5 = ½(5√(2))/d d = 9.23879532511 units. Where d is the side length of the regular triangle (purple area). Area purple is; 0.5*9.23879532511*9.23879532511sin60 = 36.95994598698 square units. Calculating Area Shaded. e = 360-135-45-67.5-60 e = 52.5° f² = 50-50cos135 f = 9.23879532511 units. g = 52.5+0.5(45) g = 75° sin(...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1916500

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AD = √(40²-20²) AD = 20√(3) units. AE = DE = ½(AD) = 10√(3) units. Angle BED is; = atan(20/(10√(3))) = atan(⅔√(3))° = 49.10660535087° Angle DAC is; = 30° (EF)² = 300+625-20√(3)*25cos30 EF = 5√(7) units. (5√(7)/sin30) = (25/sina) a = 70.89339464913° a is angle DEF. Therefore required angle is; Angle BED + Angle DEF = 49.10660535087+70.89339464913 = 120°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1916431

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Let AD = 2AB. For AB = 2 units, AD = BC = 4 units. Angle APD is; = atan(4)° PD = √(4²+1)  PD = √(17) units. Considering similar triangles APD and QCD ratios. 1 = a √(17) = 2 Crossing Multiply. a = (2/√(17)) units. a = 0.48507125007 units. a is DQ. PQ = √(17)-0.48507125007  PQ = 3.63803437554 units. Angle BPQ = 180-atan(4) = 104.03624346793° Calculating QB. Notice, BC is 4 units. (QB)² = 3.63803437554²+1-2*3.63803437554cos104.03624346793 QB = 4 unit. It implies;  BC = QB Proved....

We are smart and clever!

https://onlineedumath.com/update/we-are-smart-and-clever/1916271

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Our certified, experienced, resilient, productive and committed Math Educators will be glad to be at your service. We make teaching and learning Mathematics fun for learners, we expose the simplicity and beauty of Mathematics to our learners, and mentor them to appreciate and love Mathematics. Communicate us to mentor/educate you Mathematics better online at affordable tuition. We are smart and clever.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1916258

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Calculating R. (5+R)² = (10-R)²+(5-R)² 25+10R+R² = 100-20R+R²+25-10R+R² 20R = 100-20R+R² R²-40R+100 = 0 Therefore; R ≠ 20+10√(3) units. R = 20-10√(3) units. R = 2.67949192431 units. Calculating r. (5-r)² = r²+a² 25-10r+r² = r²+a² a = √(25-10r) units. Therefore; (5+r)² = (5-r)²+ (5+√(25-10r))² 25+10r+r² = 25-10r+r²+25+10√(25-10r)+25-10r 20r = 25+10√(25-10r)+25-10r 30r-50 = 10√(25-10r) (3r-5)² = 25-10r 9r²-30r+25 = 25-10r 9r² = 20r 9r = 20 r = ⅑(20) units. r = 2.22 units. It implies; r ÷...
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