Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1973990

a = 180-75-35 a = 70° (12/sin70) = (b/sin35) b = 7.32464753257 cm. c =2*35 c =70° Calculating R, radius of the ascribed bigger circle. d = ½(180-70) d = 55° R is; (R/sin55) = (7.32464753257/sin70) R = 6.38506663486 cm. Calculating r, radius of the inscribed big circle. (r/tan37.5) + (r/tan35) = 7.32464753257 r = 2.68167200695 cm. (a) Radius Small ÷ Radius Large to 2 decimal places is; 2.68167200695 ÷ 6.38506663486 = 0.41999123272 ≈ 0.42 (b) Calculating Area Orange in cm² to 1 d...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1973952

Let the radius of the inscribed circle be 2 units. Calculating AB. tanx = 2/a a = 2/tanx AB = 2+a AB = 2+(2/tanx) AB = (2tanx+2)/tanx Calculating CD. tanx = b/2 b = 2tanx CD = 2+b CD = 2+2tanx It implies; AB ÷ CD is; ((2tanx+2)/tanx)/(2+2tanx) = (2tanx+2)/((tanx)(2tanx+2)) = 1/tanx

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1972329

Let the two congruent lengths be 1 unit. (1/sin66) = (a/sin96) a = 1.08863974644 units. a is BC. (1.08863974644/sinb) = (1/sin48) b = 54° c = 180-54 c = 126° c is angle BDC. It implies, x the required angle, angle DCB is; 180-48-126 = 6°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1971670

Notice; AB = 25 units. cos30 = 8√(3)/a a = 16 units. tan30 = b/(8√(3)) b = 8 units. Let the side of the regular hexagon be c. c = ¼(AC) Calculating AC. ((AC)/sin46) = (25/sin74) AC = 18.70821971213 units. c = ¼(18.70821971213) c = 4.67705492803 units. Calculating r, radius of the circle. (r/sin16) = (25/sin148) r = 13.00374294827 units. Calculating Shaded Area. cos30 = d/4.67705492803 d = 4.05044838257 units. e = 2d e = 8.10089676514 units. tan16 = f/(0.5*25) f = 3.58431732199 uni...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1970168

Let the square side be 1 unit. (1/sin150) = (a/sin15) a = 0.51763809021 units. a is AE = BE. b = 90-15 b = 75° b is angle EBC. (CE)² = 0.51763809021²+1-2*0.51763809021cos75 CE = 1 unit. It implies; x, angle ECB is; 180-(2*75) = 180-150 = 30°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1970098

Let the hypotenuse of the inscribed triangle be 10 units. Calculating the radius of the ascribed half circle. Let it be a. 2a² = 10² a² = 50 a = 5√(2) units. a = 7.07106781187 units. Calculating r. (5√(2)-r)² = r²+5² 50-10√(2)r+r² = r²+25 25 = 10√(2)r 5 = 2√(2)r r = ¼(5√(2)) units. r = 1.76776695297 units. Calculating R. sin30 = a/10 a = 5 units. cos30 = b/10 b = 5√(3) units. Where a and b area adjacent base and adjacent height of the inscribed right-angled triangle with hypotenuse 10...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1969982

Notice, triangle ABC is equilateral. BD = √(3)+2-1 BD = (√(3)+1) units. BD = 2.73205080757 units. BC = 3.73205080757 units. (CD)² = 2.73205080757²+3.73205080757²-2*2.73205080757*3.73205080757cos60 CD = 3.34606521495 units. (2.73205080757/siny) = (3.34606521495/sin60) y = 45° Calculating x. It implies; x+y = 60 x = 60-y And y is 45° Therefore; x = 60-45 x = 15°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1969883

Let the radius of the three congruent inscribed circles be r. Calculating r. (1+r)² = (1-r)²+(1-3r)² 1+2r+r² = 1-2r+r²+1-6r+9r² 1+2r+r² = 10r²-8r+2 9r²-10r+1 = 0 Therefore, solving the quadratic equation. r ≠ 1 unit. r = ⅑ units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1969678

(AC)² = 2a² AC = √(2)a units. b² = 3²+4² b = 5 units. c = √(16-a²) units. d = a-c d = (a-√(16-a²)) units. Therefore, considering similar triangle ratios. a/4 = d/3. And d is (a-√(16-a²)). Therefore; a/4 = (a-√(16-a²))/3 Cross Multiply. 3a = 4a-4√(16-a²) 4√(16-a²) = a 16(16-a²) = a² 256-16a² = a² 256 = 17a² a² = 256/17 a = 16/√(17) a = (16√(17))/17 units. a = 3.88057000058 units Where a is a side length of the square ABCD.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1968584

4 = 18-18cosa a = acos(7/9) a = 38.94244126898° a is angle BOC. b = 90-38.94244126898 b = 51.05755873102° b is angle COD. d = 0.5(180-38.94244126898) d = 70.52877936551° d is angle OCB = angle OBC. e = 90-70.52877936551 e = 19.47122063449° e is angle OCD. f = 19.47122063449+51.05755873102 f = 70.52877936551° f is angle ADC. g = 0.5*38.94244126898+51.05755873102 g = 70.52877936551° sin70.52877936551 = 1/h h = 1.06066017178 units. h is OD. tan(0.5*70.528779365...

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