Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2774023

Let the two equal lengths be 1 unit each. a² = 2(1²) a = √(2) units. b = 180-60-45 b = 75° (√(2)/sin45) = (c/sin75) c = 1.9318516526 units. d = c-a d = 0.5176380902 units. e = 180-45 e = 135° It implies; f² = 0.5176380902²+1.9318516526-2*0.5176380902*1.9318516526cos135 f = √(2) units . Calculating the required angle, x. (√(2)/sin135) = (1/sinx) x = 30°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2774015

Sir Mike Ambrose is the author of the question. Please, move the above question left/right one time to review the solution. Thank you. Let the side length of the inscribed square be 1 unit. Therefore; The required angle, theta is; 45.896821403° ≈ 45.9° to 1 d. p.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2772866

Sir Mike Ambrose is the author of the question. Calculating Area Purple. Let the side length of the square be x. Calculating x. 2(one-sixth x²) = Area Orange 2(1/6)x² = 96 x² = 288 x = 12√(2) cm. Therefore; Area purple is; Area triangle with height 12√(2) cm and base 6√(2) cm - Area triangle with two side 6√(2) cm and 10.5409255339 cm, and angle atan(0.5)° = (½*12√(2)*6√(2)) - (½*6√(2)*10.5409255339sin(atan(0.5)) = 72 - 20 = 52 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2772527

(0.5*(8+3)(8+4)sina)-(0.5*3*8sina) = 36 Calculating angle a. 66sina-12sina = 36 11sina-2sina = 6 9sina = 6 sina = 2/3 a = asin(2/3) a = 41.8103148958° Therefore, the required area is; 0.5*3*8sin(asin(2/3)) = 0.5*3*8*(2/3) = 8 square units. Or It is; 0.5*12*11sin(asin(2/3))-36 = 66*(2/3)-36 = 44-36 = 8 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2772524

Let the side length of the ascribed regular pentagon be 2 units. a = ⅕*180(5-2) a = 108° a is the single interior angle of the ascribed regular pentagon. b² = 2²+1²-2*1*2cos108 b = 2.497212041 units. (2.497212041/sin108) = (1/sinc) c = 22.3861775592° d = 180-108-c d = 49.6138224408° e = 180-d e = 130.3861775592° f = d-c f = 27.2276448816° g = 180-2d g = 80.7723551184° h = 180-g-f h = 72° (2.497212041/sin72) = (j/sin27.2276448816) j = 1.2013397144 units....

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2772515

R is; R² = 2(R-2)² R² = 2(R²-4R+4) R² = 2R²-8R+8 R²-8R+8 = 0 (R-4)² = -8+16 (R-4)² = 8 R = 4±2√(2) Therefore; R ≠ 4-2√(2) cm. R = 4+2√(2) cm.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2772375

a² = 11*11+10*10-2*11*10cos50 a = 8.92113926968 cm. (8.92113926968/sin50) = (10/sinb) b = 59.16920318624° c = 180-50-59.16920318624 c = 70.83079681376° d = 0.5a d = 4.46056963484 cm. e² = 4.46056963484²+5.5²-2*4.46056963484*5.5cos59.16920318624 e = 5 cm. (5/sin59.16920318624) = (5.5/sinf) f = 70.83079681376° g² = 4.46056963484²+5²-2*4.46056963484*5cos70.83079681376 g = 5.5 cm. (5.5/sin70.83079681376) = (5/sinh) h = 59.16920318624° i = 180-59.16920318624-70.83...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2771965

Calculating the area of the ascribed right-angled triangle. a = (y+2) units. a is the adjacent base of the right-angled triangle. b = (z+2) units. b is the adjacent height of the right-angled triangle. 2 units is the radius of the inscribed circle. c = (y+z) units. c is the hypotenuse of the right-angled triangle. a²+b² = c² (y+2)²+(z+2)² = (y+z)² --- (1). tanx = 2/z --- (2). tanx = 2/(y+2) --- (3). Equating (2) and (3). 2/z =2/(y+2) z = y+2 --- (4). Substitu...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2771184

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/2771080

tana = 3/4 a = atan(3/4) b² = 3²+4² b = 5 units. cos(atan(3/4)) = 1/c c = 1.25 units. c = (5/4) units. sin(atan(3/4)) = 1/d d = 1.6666666667 units. d = (5/3) units. e = b-c-d e = 5-(5/4)-(5/3) e = (60-15-20)/12 e = 25/12 units. e is AB. sin(atan(3/4)) = f/(25/12) f = (5/4) units. f is BC. cos(atan(3/4)) = g/(25/12) g = (5/3) units. g is AC. Therefore, area triangle ABC (yellow area) is; ½*fg = ½*(5/4)*(5/3) = (25/24) square units. = 1.0416666667 square...

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