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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2321738

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Let the three equal lengths be a. Radius r of the semi circle is; r = 5+4 r = 9 units. 5² = b²+a²-2abcos45 25 = b²+a²-√(2)ab --- (1). 9² = b²+a²-2abcos135 81 = b²+a²+√(2)ab --- (2). Solving (1) and (2) simultaneously. Subtracting (1) from (2). 81 = b²+a²+√(2)ab -(25 = b²+a²-√(2)ab) 56 = 2√(2)ab 28 = √(2)ab a = 28/√(2)b a = 14√(2)/b --- (3). Substituting (3) in (1) to get b. 25 = b²+(14√(2)/b)²-√(2)(14√(2)/b)b 25 = b²+(392/b²)-28 53 = b²+(392/b²) 53 = (b⁴+392)/b² 53b² = b⁴+392 b⁴-53b²+3...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2321437

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Let the three equal lengths be 1 unit each. a = 180-15-45 a = 120° b = 180-120 b = 60° It implies; (c/sin45) = (1/sin15) c = 2.7320508076 units d² = 2.7320508076²+2²-2*2*2.7320508076cos60 d = 2.4494897428 units. Therefore, the required angle x is; (2.4494897428/sin60) = (2.7320508076/sinx) sinx = 0.9659258263 x = asin(0.9659258263) x = 75°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2320358

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a = 180-50-30 a = 100° a is angle AED. Let DE = 1 unit. (1/sin30) = (b/sin100) b = 1.969615506 units. b is AD. c = 180-80-50 c = 50° c is angle CAD. (1.969615506/sin80) = (d/sin50) d = 1.5320888862 units. d is CD. CE = 1.5320888862-1 CE = 0.5320888862 units. e = 180-50 e = 130° e is angle ADB. f² = 0.5320888862²+1.969615506²-2*1.969615506*0.5320888862cos130 f = 2.3472963553 units. f is AB. Therefore, calculating the required angle, angle BAD (angle x). (2.3472963553/sin130) = (1.96961...

We are smart and clever!

https://onlineedumath.com/update/we-are-smart-and-clever/2320203

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Online Edumath Educators and Learners are Super Smart and Amazingly, Very Clever. Communicate us to mentor/teach/educate your child/children Mathematics online at affordable tuition, helping them become a lover of Mathematics, and also, a Mathematician. Call/WhatsApp: +2349076614992 or +2348152827445 Email: [email protected] [email protected] [email protected] Facebook Page: Online Edumath We are super smart and amazingly, very clever!

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2320171

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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 = 180-59.16920318624-70.83079681376 i = 50° (j/tan...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2319903

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Let the length of the rectangle be (a+3) cm. Let the width of the rectangle be (b+2) cm. Therefore, Calculating a and b. Notice! The area of the two inscribed triangles and quadrilateral are equal. It implies; ½(a(2+b)) = ½(b(3+a)) 2a+ab = 3b+ab 2a = 3b a = ½(3b) cm. Length = (½(3b)+3) Length = ½(3b+6) cm. Therefore; ½(3b+6)(b+2) = 3*½(b*½(3b+6)) ½(3b²+6b+6b+12) = ¼(9b²+18b) 3b²+12b+12 = ½(9b²+18b) 6b²+24b+24 = 9b²+18b 3b²-6b-24 = 0 b²-2b-8 = 0 Resolving the above quadratic equation usi...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2319845

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a² = 85²-68² a = 51 units. a is the height of the ascribed triangle. Calculating r, radius of the inscribed circle. 85r+51r+68r = 51*68 204r = 3468 r = 17 units. b = 51-17 b = 34 units. tanc = (68/51) c = atan(68/51) Therefore, shaded area yellow is; ½(68*51)-½(34*34)sin(atan(68/51) = 1734-462.4 = 1271.6 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2318812

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Notice! Length 3 units and length 6 units are both tangent to the green inscribed circle. It implies; Radius r, of the ascribed semi circle is; r = ½(3+6) r = ½(9) r = 4.5 units. Where 9 units is the diameter of the ascribed semi circle. 9² = 5²+6²-2*5*6cosa 81 = 61-60cosa 60cosa = 61-81 cosa = (-20/60) a = acos(-1/3) a = 109.4712206345° a is angle ADB. b = 180-a b = 180-109.4712206345 b = 70.5287793655° b is angle BDC. Let  length CD, the required length be x. It implies; cos70.5287793...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2315557

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Let the side length of the regular pentagon be a. b = ⅕(180(5-2)) b = 108° b is the single interior angle of the regular pentagon. c = 108-90 c = 18° d = (180-2(18)) d = (180-36) d = 144° e = ½(f) e = ½(144) e = 72° notice! Radius of the circle is 1 unit. It implies; sin72 = 0.5a/1 0.5a = sin72 a = 2sin72 units. a is the side length of the regular pentagon which is also the side length of the inscribed square. It implies; Area square exactly is; a² = (2sin72)² = 4sin²(72) square units...

We are smart and clever!

https://onlineedumath.com/update/we-are-smart-and-clever/2315187

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Online Edumath Educators and Learners are Super Smart and Amazingly, Very Clever. Communicate us to mentor/teach/educate your child/children Mathematics online at affordable tuition, helping them become a lover of Mathematics, and also, a Mathematician. Call/WhatsApp: +2349076614992 or +2348152827445 Email: [email protected] [email protected] [email protected] Facebook Page: Online Edumath We are super smart and amazingly, very clever!
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