NO!!!
Sine40 = Cosine50 = 0.64278...
Sin30 = Cos 60 = 0.5
Sin 0 = Cos 90 = 0
Sin 90 = Cos 0 = 1
From Latin the word 'Sine' means 'curve'. Cosine means 'Complimentary Sine' or Complimentary curve'.
The 'awkward' decimal numbers are based on a the radius of a circle, that radius being the hypotenuse of a right- triangle based at the circles origin/centre.
It is:- sin(40) = 0.6427876097
To determine which is greater between 50% of 40 and 40% of 50, we can calculate the actual values. 50% of 40 is 0.50 x 40 = 20. 40% of 50 is 0.40 x 50 = 20. Therefore, both values are equal at 20.
sin(40o) = 0.6428
135
A nonagon is a nine-sided polygon. The area of a regular polygon is:A = (n/4)(s^2)[cos (180° /n)]/[sin (180° /n)]So the area of the nonagon with side 16 (144/9) is:A = (9/4)(16^2)[[cos(180°/9)]/[sin (180°/9)]]= (2.25)(256)[(cos 20°)/(sin 20°)]A ≈ 1,582.55 ft^2 If you don't know the formula of the area of a polygon, you can find its area by multiplying by 9 the area of one of the 9 congruent isosceles triangles that are formed by connecting the center of the polygon with its vertices. But for this you need to find the altitude and the length of the side (which is the radius of the circumscribed circle) of that triangle such as:we know the length base which is 16 ft (144/9), the angle base which is 70 (140/2), and the vertex angle which is 40 (360°/9 or 180° - 140°). By using the Law of Sines we can find the length of r. So,r/sin 70° = 16/sin 40° multiply by sin 70° to both sides;r = (16 sin 70°)/sin 40° sin 70 = altitude/radiusaltitude = (sin 70)(radius) = (Sin 70)[(16 sin 70)/sin 40]altitude = [16(sin 70)^2]/sin 40Thus the area of this nonagon is:A = 9[(1/2)(bh)] where b = 16 and h = [16(sin 70)^2]/sin 40A = (4.5)(16) [[16(sin 70)^2]/sin 40] A ≈ 1,582.55 ft^2
Like normal expansion of brackets, along with: cos(A + B) = cos A cos B - sin A sin B sin(A + B) = sin A cos B + cos A sin B 5(cos 20 + i sin 20) × 8(cos 15 + i sin 15) = 5×8 × (cos 20 + i sin 20)(cos 15 + i sin 15) = 40(cos 20 cos 15 + i sin 15 cos 20 + i cos 15 sin 20 + i² sin 20 sin 15) = 40(cos 20 cos 15 - sin 20 cos 15 + i(sin 15 cos 20 + cos 15 sin 20)) = 40(cos(20 +15) + i sin(15 + 20)) = 40(cos 35 + i sin 35)
sine 40° = 0.642788
cos 71
Sadly, no. 20 does not equal 611, 40 does not equal 611, and just now, I can't think of any mathematical operation that could be applied to 20 and 40 with a result of 611.
20% of 40 is not greater than 40% of 20. The two are equal. 20% of 40 = 0.20 * 40 = 8 40% of 20 = 0.40 * 20 = 8
sin(angle) = opposite/hypotenuse → hypotenuse = opposite/sin(angle) opposite = rise → hypotenuse = 40ft / sin 16.5° ≈ 140.84 ft
20 and 20
62
20
No. All equilateral and equiangular triangles are acute. (All angles are equal to 60°, which is less than a right angle [90°]); however, the converse (which is what was asked) is not true.A triangle can have all three angles be less than 90°, but not be an equilateral triangle.An example is a triangle with angles of 80°, 60°, and 40°. It is scalene and acute.From the Law of Sines: a/sin(A) = b/sin(B) = c/sin(C), you can show that sin(80°) does not equal sin(60°) or sin(40°), so none of sides a, b, and c, are equal.You could have an acute isosceles triangle like: 80°, 80° and 20° angles, as another example. From the Law of Sines, you can show that two of the sides are equal, but the third side (opposite the 20° angle) is not equal to either of the other 2.
20 x 2 = 40
360