By OnlineEdumath   |  8th February, 2025
Let the length of the ascribed regular hexagon be 2 units. Therefore it area will be; (2*2*6)/(4tan(180/6)) = 6/tan30 = 6√(3) square units. The green areas is; Area rectangle with...
By OnlineEdumath   |  8th February, 2025
Let the side of the ascribed regular hexagon be 2 units. Area regular hexagon is; (2*2*6)/(4tan(180/6)) = 6/tan30 = 6√(3) square units. Area green is; Area kite with perpendicular lengths 4 un...
By OnlineEdumath   |  7th February, 2025
a² = 4²+4²-2*4*4cos150 a = 7.72740661031 cm. Where a is BE. b = angle ABE = ½(180-150) b = 15° c = angle AFE = 180-45-15 c = 120° (4/sin120) = (d/sin45) d = 3.26598632371 cm. Where d is EF. Noti...
By OnlineEdumath   |  7th February, 2025
tan15 = a/4 a = 1.07179676972 cm. Where a is AD. b² = 1.07179676972²+4² b = 4.14110472164 cm. Where b is DF. c = 4-1.07179676972 c = 2.92820323028 cm. Where c is CD. sin60 = d/2.92820323028  d =...
By OnlineEdumath   |  6th February, 2025
a = acos(3/5) Where a is angle GCA. tan(acos(3/5)) = b/3 b = 4 units. Where b is AG. c = atan(⅔) Where c is angle GBE. d = 180-acos(3/5)-atan(⅔) d = 93.17983011986° Where d is angle BEC. (6/sin9...
By OnlineEdumath   |  6th February, 2025
a = asin(1/6) b = 2a b = 2asin(1/6) Where b is angle BAC cos(2asin(1/6)) = c/6 c = ⅓(17) units. Where c is AD. d = 6-c d = 6-⅓(17) d = ⅓ units Where d is CD. sin(2asin(1/6)) = e/6 e = 1.97202659...
By OnlineEdumath   |  5th February, 2025
a = 15-11.25  a = 3.75 m. b = atan(3/3.75)° c = 90-atan(3/3.75) c = 51.34019174591°  Calculating Length CD. tan(51.34019174591) = 15/CD CD = 15/tan(51.34019174591)  CD = 12 m....
By OnlineEdumath   |  5th February, 2025
Let the square side be a. tan60 = a/b b = a/tan60 cm. Where b is BS. Calculating a. 2b+a = 6 2(a/tan60)+a = 6 a((2/tan60)+1) = 6 a = 6/((2/tan60)+1) a = 6(2√(3)-3) cm. Area Shaded...
By OnlineEdumath   |  4th February, 2025
a² = 10²+10²-2*10*10cos150 a = 19.31851652578 cm. Where a is AF which is equal AE. (19.31851652578/sin150) = (10/sinb) b = 15° Where b is angle BAF. tan15 = c/10 c = 2.67949192431 cm....
By OnlineEdumath   |  4th February, 2025
a = atan(2)° b = 180-2a b = (180-2atan(2))° Where b is angle FBP. Notice; CQ = 5 cm. Calculating FP using sine rule. Let FP be c. (c/sin(180-2atan(2))) = (5/sin(atan(2))) c = 4...
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