Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2569978

let theta be a. tana = k/3k a = atan(⅓)° Notice! 3k = 4. b = 2a b = 2atan(⅓)° It implies; tan(2atan(⅓)) = c/4 c = 3 units. Therefore, ? the required length is; Let if be d. d = 4-c d = 4-3 d = 1 unit.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2569924

Calculating angle x (angle CBD). Let AC = 1 unit. a = 180-12-18 a = 150° a is angle ADC. (b/sin12) = (1/sin150) b = 0.4158233816 units. b is CD. c = 180-(54+12)-(48+18) c = 180-66-66 c = 48° c is angle ABC. (1/sin48) = (d/sin66) d = 1.2292966678 units. d is BC = AB. e² = 1.2292966678²+0.4158233816²-2*1.2292966678*0.4158233816cos48 e = 1 unit. Therefore, the required angle x (angle CBD) is; (1/sin48) = (0.4158233816/sinx) x = 18°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2569865

a = 180-54-48 a = 180+102 a = 78° b = 30+48 b = 78° c = 180-2*72 c = 36° Let the two congruent side lengths be 1 unit each. d² = 2-2cos36 d = 0.6180339887 units. e = 180-30-24 e = 126° f = 180-72 f = 108° (0.6180339887/sin30) = (g/sin24) g = 0.5027541398 units. (0.5027541398/sin78) = (h/sin48) h = 0.3819660113 units. Notice! h+d = 1 unit. (0.3819660113+0.6180339887 = 1) Therefore, the required angle, alpha is; Alpha = ½(180-72) Alpha = ½(108) Al...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2569825

Let AD be x. tana = 4/x --- (1). tana = x/9 --- (2). Equating (1) and (2). 4/x = x/9 Cross Multiply. x² = 36 x = 6 units. Again, x is AD, the height of the ascribed trapezoid ABCD. Therefore, area trapezoid ABCD is; ½(4+9)*6 = ½(6*13) = 3*13 = 39 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2569789

a = ½(24) a = 12 units. a is the radius of the half by circle. sinb = 8/24 b = asin(⅓)° tan(asin(⅓)) = c/12 c = 4.2426406871 units. tand = 12/4.2426406871 d = 70.5287793656° e = ½(d) e = 35.2643896828° tan35.2643896828 = r/f f = 1.4142135624r units. g = f+c g = (1.4142135624r+4.2426406871) units. h = (12-r) units. Calculating r, radius of the inscribed circle. (12-r)² = r²+(1.4142135624r+4.2426406871)² 144-24r = (1.4142135624r+4.2426406871)² 144-24r = 2r²+12r+18 2r²+36r-126 = 0 r²+18r...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2569702

a = (r-7) units. b = (r-14) units. Therefore, r, radius of the inscribed circle is; r² = a²+b² r² = (r-7)²+(r-14)² r² = r²-14r+49+r²-28r+196 r²-42r+245 = 0 Resolving the above quadratic equation via factorization approach to get r, radius of the inscribed circle. r²-35r-7r+245 = 0 r(r-35)-7(r-35) = 0 (r-35)(r-7) = 0 Therefore; r ≠ 7 units. r = 35 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2569608

a = 6 units. a is the side length of the square. Therefore; Area square is; 6*6 = 36 square units. Area inscribed shaded triangle is; ½(6*6) = 18 square units. It implies; The percentage of the square that is shaded is; (18/36)*100 = 50%

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2569584

a² = 24²+7² a = √(625) a = 25 units. b = 7+a b = 32 units. c² = 32²+24² c = √(1600) c = 40 units. d = ½(c) d = 20 units. It implies, the required length, x is; 25² = x²+20² x² = 625+400 x = √(225) x = 15 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2569151

Calculating r, radius of the inscribed half circle. a = (r-6) units. b = (r-3) units. Therefore; r² = a²+b² r² = (r-6)²+(r-3)² r² = r²-12r+36+r²-6r+9 r²-18r+45 = 0 Resolving the above quadratic equation via factorization approach to get r, radius of the inscribed half circle. r²-15r-3r+45 = 0 r(r-15)-3(r-15) = 0 (r-15)(r-3) = 0 Therefore; r ≠ 3 units. r = 15 units. Again, ris the radius of the inscribed half circle.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2569122

Let the side length of the regular hexagon be 1 unit. Therefore, area A (area inscribed square) is; 1² = 1 square unit. Calculating area B. a² = 1²+(½)² a² = 1+¼ a = √(5/4) a = ½√(5) units. a = 1.1180339887 units. a is the radius of the inscribed arc. b = ⅙*180(6-2) b = 120° b is the single interior angle of the regular hexagon. It implies; (½√(5)/sin120) = (1/sinc) c = 50.7684795195° d = 60-c d = 9.2315204805° Area B is; 0.5*1.1180339887sin9.2315204805...

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