Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2794161

a = ¼(60) a= 15 units. b² = 40²+15² b = 5√(73) units. b = 42.7200187266 units. c = ½(b) c = 2.5√(73) units. c = 21.3600093633 units. c is each of the congruent lengths. d = 60-15 d = 45 units. tane = 15/40 e = atan(3/8)° f = 90-e f = atan(8/3)° Therefore, the required length, ? is; Let it be g. g² = 21.3600093633²+45²-2*21.3600093633*45cos(atan(8/3)) g = 42.5 units. Again, g is the required length.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2793964

a = 180-54-30-18 a = 180-102 a = 78° a is angle ABC. b = 78+54 b = 132° b is angle BAD. c = 180-b-18 c = 180-150 c = 30° c is angle ADB. Let AD = 1 unit. (1/sin18) = (d/sin30) d = 1.6180339887 units. d is AB. (1.6180339887/sin54) = (e/sin(18+30)) e = 1.486289651 units. e is AC. f² = 1²+1.486289651²-2*1*1.486289651cos54 f = 1.2090569266 units. f is CD. Therefore, the required angle, ? is; Let it be g. (1.2090569266/sin54) = (1/sing) g = 42° Again, g i...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2793942

Calculating h. tan60 = h/a a = ⅓(√(3)h) units. tan30 = h/b b = √(3)h units. It implies; a + b = 100 ⅓(√(3)h)+√(3)h = 100 √(3)h+3√(3)h = 300 4√(3)h = 300 √(3)h = 75 h = ⅓(75√(3)) units. h = 25√(3) units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2793317

Let a be the radius of the half circle. a-x = 4 x = a-4 --- (1). b = (a-2x) units. It implies; a² = 4²+(a-2x)² a² = 16+a²-4ax+4x² 4ax = 16+4x² --- (2). Substituting (1) in (2) to get a, radius of the half circle. 4a(a-4) = 16+4(a-4)² 4a²-16a = 16+4(a²-8a+16) 4a²-16a = 16+4a²-32a+64 16a = 80 a = 5 units. Again, a is the radius of the half circle. Recall. a-x = 4 And a = 5 units. 5-x = 4 x = 1 unit. c = 2x c = 2 units. d = 8-c d = 6 units. It implie...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2792535

Radius of the quarter circle is (2+3) = 5 units. Let a be the side length of the inscribed square. b²+3² = a² b = √(a²-9) units. c = 3+b c = (3+√(a²-9)) units. Calculating a. (3+√(a²-9))²+3² = 5² 9+6√(a²-9)+a²-9+9 = 25 9+6√(a²-9)+a² = 25 6√(a²-9) = 16-a² 36(a²-9) = (16-a²)² 36a²-324 = 256-32a²+a⁴ a⁴-68a²+580 = 0 It implies; a = √(10) units a = 3.1622776602 units. Again, a is the side length of the inscribed square. Recall. b = √(a²-9) units And a = √(10)...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2792503

Sir Mike Ambrose is the author of the question. Let the ascribed quarter circle radius be 4 units. Therefore; Area blue is; Area triangle with two side 1 unit and 3 units, and angle 2asin(1/3)° = ½*3sin(2asin(1/3)) = 0.94280904158 space units. Area red is; Area triangle with two side 1 unit and 3 units, and angle (180-2asin(1/3))° + Area triangle with two side 2 units and √(44/3) units, and angle 100.02498786° = ½*3sin(180-2asin(1/3)) + ½*2*√(44/3)sin100.02498786 = 0.942809...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2791220

Let a be the radius of the circle. 10² = 2a²+2a²cosd 50 = a²+a²cosd --- (1). 10² = (2a)²+e² e = √(100-4a²) units. √(100-4a²)² = 2a²-2a²cosd 100-2a² = 2a²-2a²cosd 2a²cosd = 4a²-100 cosd = (2a²-50)/a² --- (2). Substituting (2) in (1). 50 = a²+a²(2a²-50)/a² 100 = 3a² a² = ⅓(100) a = ⅓(10√(3)) units. a = 5.7735026919 units. Again, a is the radius of the circle and also the width of the blue area. cose = 5/a cose = 5/⅓(10√(3)) cose = 15/(10√(3)) cose = 3/(2√(3)) cos...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2790874

Sir Mike Ambrose is the author of the question. Let the side of the regular pentagon be 2 units. cos36 = a/2 a = 1.61803398875 units. Where a is the radius of the semi circle. Area semi-circle is; ½(1.61803398875)²π = 4.11239817295 square units. Calculating area S. b = 108-36 b = 72° c = 1.61803398875-1 c = 0.61803398875 units. tan72 = d/0.61803398875 d = 1.90211303259 units. e = atan(1.90211303259)° e = 62.26769855229° (f/sin62.26769855229) = (1.61803398875/s...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2790681

Let a be the radius of the semi circle. b²+24² = (2a)² b = √(4a²-576) units. b is BC. c = b-BP c = (√(4a²-576)-7) units. c is CP. It implies; Calculating a. 2a - (√(4a²-576)-7) √(4a²-576) - a Cross Multiply. 2a² = √(4a²-576)*(√(4a²-576)-7) 2a² = 4a²-576-7√(4a²-576) 7√(4a²-576) = 2a²-576 49(4a²-576) = (2a²-576)² 196a²-28224 = 4a⁴-2304a²+331776 4a⁴-2500a²+360000 = 0 Therefore; a = 20 units. Again, a is the radius of the half circle. Recall. b = √(4a²-...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2790625

Calculating the required angle, theta. Let it be a. Let the side length of the 4 congruent squares be 1 unit each. Therefore. tanb = 1/3 b = atan(⅓)° tanc = 1/(½) c = atan(2)° d = 180-c d = (90+atan(½))° Therefore; a+b+d = 180 a = 180-atan(⅓)-(90+atan(½)) a = 90-atan(⅓)-atan(½) a = (atan(2)-atan(⅓))° a = 45°

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