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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1894594

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Let the two congruent angles be a. Let the base of the ascribed triangle be b. tana = 8/b b = 8/(tana) ----- (1). tan2a = 18/b b = 18/(tan2a) ----- (2). Equating (1) and (2). 8/(tana) = 18/(tan2a) And tan2a = 2tana/(1-tan²a) Therefore; 8/(tana) = 18/(2tana/(1-tan²a)) Let p be tana. 8/p = 18/(2p/(1-p²)) 8/p = 18(1-p²)/2p Cross Multiply. 16p = 18p(1-p²) 16p = 18p-18p³ 16 = 18-18p² 18p² =2 p² = ⅑ p = ⅓ And p = tana tana = ⅓ a = atan(⅓) a = 18.43494882292° Calculating b, base of the t...

We are smart and clever!

https://onlineedumath.com/update/we-are-smart-and-clever/1893230

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Screenshot of today's teaching and learning activities with super smart and clever Adesuwa (year 5) and Oghosa (year 7). Educators: Mr. Oyoma, Mr. Daniel We are smart and clever.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1892805

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Using 1 unit base. tan70 = 1/a a = 1/tan70 a = 0.36397023427 units. tan80 = 1/b b = 1/tan80 b = 0.17632698071 units. tan40 = c/a c = 0.36397023427tan40 c = 0.30540728934 units. Therefore, the required angle ? is; Let it be d. tand = (0.17632698071/0.30540728934) d = atan(0.17632698071/0.30540728934) d = 29.99999999959° d = 30°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1892765

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Let the radius of the ascribed quarter circle be a. a² = 2(a-1)² a² = 2(a²-2a+1) a² = 2a²-4a+2 a²-4a+2 = 0 (a-2)² = -2+4 a = 2±√(2) a ≠ 2-√(2) a = 2+√(2) units. a = 3.41421356237 units. Where a is AB = BC. BQ = AB-AQ BQ = 2+√(2)-2 BQ = √(2) units. Area Blue, Area Triangle CBQ is; ½*√(2)*(2+√(2)) = ½(2√(2)+2) = (√(2)+1) square units. = 2.41421356237 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1892231

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Let DE = BF = a. Let EG = GH = HF = b. (3b)² = 12²+(12-2a) 9b² = 144+144-48a+4a² b² = ⅑(288-48a+4a²) ----- (1). CF = √(144+a²) (CG)² = (12-a)²-b² CG = √((12-a)²-b²) Therefore; (144+a²) = 4b²+((12-a)²-b²) 144+a² = 4b²+144-24a+a²-b² 24a = 3b² b² = 8a ----- (2). Equating (1) and (2). ⅑(288-48a+4a²) = 8a 4a²-120a+288 = 0 a²-30a+72 = 0 (a-15)² = -72+225 a = 15±√(153) a ≠ 15+3√(17) a = 15-3√(17) cm. a = 2.63068312315 cm. AF = 12-2.63068312315 AF = 9.36931687685 cm. HF = √(8a) HF = √(8...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1890114

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Let r be the radius of the inscribed circle. Calculating r. tan30 = r/(6-r) √(3)r = 6-r r = 6/(√(3)+1) cm. r = 3(√(3)-1) cm. sin30 = r/a ½ = 3(√(3)-1)/a a = 6(√(3)-1) cm. Where a is BO, O is the center of the inscribed circle. BP = 6(√(3)-1)-r BP = 6√(3)-6-3√(3)+3 BP = 3√(3)-3 BP = 3(√(3)-1) cm. cos30 = 6/(BH) BH = 12/√(3) BH = 4√(3) cm. PH = BH-BP PH = 4√(3)-3√(3)+3 PH = (√(3)+3) cm. tan30 = (DH)/6 DH = 6/√(3) DH = 2√(3) cm  HR = 2√(3)-3√(3)+3 HR = (3-√(3)) cm. (PR)² = (√(3)+3)²+(3-√...

We are smart and clever!

https://onlineedumath.com/update/we-are-smart-and-clever/1886580

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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. We are smart and clever.

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/1886431

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(AM)² = 15²+8²-2*15*8cos60 AM = √(225+64-120) AM = √(105+64) AM = √(169) AM = 13 units. (13/sin60) = (8/sina) a = 32.20422750397° Where a is angle BAM. b = 90-32.20422750397  b = 57.79577249603° Where b is angle DAM. Therefore, calculating DM. (DM)² = 13²+11²-2*13*11cos57.79577249603 DM = 11.72943003449 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1886366

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Let AB = BC = CD = 1 unit. AC = √(1²+1²) AC = √(2) units. (AD)² = 3-2√(2)cos105 AD = 1.93185165258 units. Therefore; (1.93185165258/sin105) = (√(2)/sin?) ? = 45° Where ? is angle ADC.
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