6² = 4²+6²-2*6*4cosa
48cosa = 52-36
cosa = 16/48
a = acos(⅓)
a = 70.5287793655°
b = 180-70.5287793655
b = 109.4712206345°
c = 180-2(70.5287793655)
c = 38.942441269°
d = ½(180-70.52877...
a = 180-60
a = 120°
b = 180-120-45
b = 15°
(1/sin15) = (c/sin45)
c = 2.7320508076 units.
d² = 2.7320508076²+2²-2*2*2.7320508076cos60
d = 2.4494897428 units.
It implies;
Calculating angle x.
(2...
Equation of the curve is;
y = ¼(x²)+4
Shaded area exactly in square units is;
Area triangle with height 8 unit and base 4 unit - Area sector with radius 4√(5) unit and angle (2atan(0.5))° + Area...
Area orange exactly in square units is;
Area triangle with height ⅓(10√(2)) unit and base 5sin45 unit - Area triangle with height (8/3) unit and base 2√(2)sin45 unit - Area triangle with height ⅔√(...
Sir Mike Ambrose is the author of the question.
Let the side of the square be 2 units.
Therefore;
a = 3√(5)/5 unit
b = 2√(5)/5 unit
c = √(5)/5 unit
d = 2√(5)/5 unit
It implies;
a² = b² + c² + d²...
Please, move the above question left or right one time to review the solution.
Thank you.
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Let the side length of the regular Pentagon be 1 unit.
Therefore Area Pentagon is;
(5*1)÷(4tan(180/5))
= 5/(4tan36) space unit.
= 1.72047740059 space unit.
Area star is;
5(area triangle with heig...
At x = 0, y = 2.
At y = 16, x = 1.
Gradient of the curve at x = 1 or at the tangent is;
dy/dx = 3In2*2^(3x+1)
For x = 1
dy/dx = 48In2
Therefore coordinate Q is (x¹, 0).
Calculating x¹.
(0-16)/(x¹...
Let the side length of the be 2 unit.
Therefore;
BC = 2 unit.
It implies;
Area orange is;
½(1*2) = 1 square unit.
Area green is;
½(1*2) + ½(2/√(5))(4/√(5))
= 1 + (4/5)
= (9/5)...
