Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1866128

a² = 6²+12² a = 6√(5) cm. Where a is CE = EF. b² = 2(a²) b² = 2(6√(5))² b² = 360 b = 6√(10) cm. Where b is CF. Angle DCE = atan(½)° Angle ECF = 45° Angle FCM = 90-45-atan(½) Angle FCM = 18.43494882292° tan18.43494882292 = c/12 c = 4 cm. Angle AEF = 90-atan2 Angle AEF = atan(½)° (AF)² = 36+180-2*6*6√(5)cos(atan(½)) AF = 6√(2) cm. (6/sind) = (6√(2)/sin(atan(0.5))) d = 18.43494882292° Where d is angle AFE. Angle CFM = 180-45-18.43494882292 Angle CGM = 116.56505117708° Angle CMF = 180-1...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1865815

Let CE = 1 unit. (1/sin20) = (a/sin40) a = 1.87938524157 units. Where a is AE. Notice; b = a-1 b = 1.87938524157-1 b = 0.87938524157 units. Where b is AD. Angle AEC = 180-20+40 Angle AEC = 120° (c/sin120) = (1/sin20) c = 2.53208888624 units. Where c is AC = BC. BE = 2.53208888624-1 BE = 1.53208888624 units. Angle BED = 180-120 = 60° (BD)² = 1.53208888624²+1-2*2.53208888624cos60 BD = 1.34729635534 units. Calculating Angle BDE (alpha, the required angle)....

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1865498

a = ½(180-150) a = 15° b = 60-15 b = 45° c² = 2(6²) c = 6√(2) cm. d² = 72+12²-144√(2)cos135 d= 6√(10) cm. (6√(10)/sin135) = (12/sine) e = 26.56505117708° f = 180-135-26.56505117708 f = 18.43494882292° g = 90-18.43494882292 g = 71.56505117708° tan71.56505117708 = 12/h h = 4 cm. tan71.56505117708 = j/2 j = 6 cm. It implies the height and base of the shaded region is 6 cm and 4 cm respectively. Therefore Area Shaded Region (triangle) is; ½*4*6 = ½(24) = 12 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1864230

(6/sin(180-45-atan(2))) = (a/sin45) a = 4.472135955 cm. b = √(12²+6²)+a b = 17.88854382 cm. c² = 4.472135955²+6²-12*4.472135955cos(180-atan2) c = 8.94427191 cm. Where c is CF. (8.94427191/sin(180-atan2)) = (6/Sind) d = 36.86989764584° d is angle CFD. Angle BCF is; 180-(180-atan2)-36.86989764584° = atan2-36.86989764584° = 26.56505117708° Angle ECF is; 90-26.56505117708 = 63.43494882292° Notice; atan(½)+63.43494882292 = 90° It implies; sin(atan(½)) = e/12 e = 5.366563146 cm. cos(ata...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1864150

Area Blue Triangle is; ½*10*10sin(90-2tan(½)) = 50sin(90-2tan(½)) = 50(⅗) = 30 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1863346

tan60 = a/5 a = 5√(3) cm. Where a is AD. b = ½(a) b = ½(5√(3)) cm c = atan((½(5√(3))/5) c = atan(½√(3)) Where c is angle AGB d = atan(5√(3)/5) d = atan(√(3)) Where d is angle CAG. e = 180-c-d e = 180-atan(½√(3))-atan(√(3)) e = 79.10660535087° Where e is angle AFG. Therefore; (5/sin79.10660535087) = (f/sin(atan(√(3)))) f = 4.40958551844 cm. Where f is FG. It implies; Shaded Area is; Area triangle with height 4.40958551844 cm and base 5sin(atan(0.5√(3))) cm. = 0.5*4.40958551844*5sin(...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1861855

tanx = 4/(40-4) tanx = (1/9) x = atan(1/9) y = 2x y = 2atan(1/9) Therefore; tan(2atan(1/9)) = h/40 (9/40) = (h/40) h = 9 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1861301

Let AB = x Let r be the radius of the semi circle. It implies; OC = x-r y = √(r²-(x-r)²) y = √(r²-(x²-2xr+r²)) y = √(2xr-x²) Therefore; 6²+r² = (x-r)²+(2+√(2xr-x²))² 6²+r² = x²-2xr+r²+4+4√(2xr-x²)+2xr-x² 6²+r² = r²+4+4√(2xr-x²) 32 = 4√(2xr-x²) 8 = √(2xr-x²) 64 = 2xr-x² 64+x² = 2xr r = (64+x²)/(2x) -------- (1) z = √((x+r)²-r²) z = √(x²+2xr+r²-r²) z = √(x²+2xr) It implies; (√(x²+2xr)+6)² = (2x)²+(2+√(2xr-x²))² x²+2xr+12√(x²+2xr)+36 = 4x²+4+4√(2xr-x²)+2xr-x² 2x²+12√(x²+2xr)+32 = 4x²+4√(...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1860074

r² = x²+√(8)² x = √(r²-8) units. Calculating r, radius of the semi circle. 3² = x²+(√(8)-(r/2))² 9 = √(r²-8)²+8-2√(2)r+(r²/4) 9 = r²-8+8-2√(2)r+(r²/4) 36 = 4r²-8√(2)r+r² 5r²-8√(2)r-36 = 0 Solving the quadratic equation using completing the square approach. r²-((8√(2)r)/5)-(36/5) = 0 (r-(4√(2)/5))² = (36/5)+(4√(2)/5)² (r-(4√(2)/5))² = (36/5)+(32/25) (r-(4√(2)/5))² = (180+32)/25 r-(4√(2)/5) = ±√(212/25) r = (4√(2)/5)±(2√(53)/5) r = ⅕(4√(2)±2√(53)) r ≠ ⅕(4√(2)-2√(53)) units. r = ⅕(4√(2)+2√(5...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1859501

Please, move a step right to review the analyzed figure. Considering similar triangle ABG ratios. AG = x 12 = 12-x It implies; AG = 12x/(12-x) Notice; Angle ABG = Angle EFI, therefore considering similar triangle ratios of the both triangles. (6-x) = 12x/(12-x) (12-x) = 12 It implies; 12(6-x) = (12-x)(12x/(12-x)) 72-12x = 12x 24x = 72 x = 3 cm. Therefore; AG = 4 cm. DG = 8 cm. Angle ABG = Angle EFI = atan(⅓)° (CG)² = 12²+8² CG = 4√(13) cm. Angle FEI = atan(3)° Angle CEH = (180-...

You Searched for "delta state office ekrejeta house 2 zenith bank abraka delta state nigeria"

2136 result found.

Mathematics Question and Solution

Mathematics Question and Solution

Mathematics Question and Solution

Mathematics Question and Solution

Mathematics Question and Solution

Mathematics Question and Solution

Mathematics Question and Solution

Mathematics Question and Solution

Mathematics Question and Solution

Mathematics Question and Solution

OnlineEdumath

Socials

Useful Tags

Quick Links

Reach Us

Safety and Abuse Reporting

Thanks for being awesome!

Are you sure you want to report abuse against this website?

Search

You Searched for "delta state office ekrejeta house 2 zenith bank abraka delta state nigeria"

2136 result found.

Subscribe to our mailing list for latest updates.

Safety and Abuse Reporting

Thanks for being awesome!

Are you sure you want to report abuse against this website?