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tan(9) + tan(81) - tan(27) - tan(63) = 4

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Q: Tan 9 plus tan 81 -tan 27-tan 63?
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Tan 9-tan 27-tan 63 plus tan81 equals 4?

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How tan9-tan27-tan63 tan81 equals 4?

tan(9) + tan(81) = sin(9)/cos(9) + sin(81)/cos(81)= {sin(9)*cos(81) + sin(81)*cos(9)} / {cos(9)*cos(81)} = 1/2*{sin(-72) + sin(90)} + 1/2*{sin(72) + sin(90)} / 1/2*{cos(-72) + cos(90)} = 1/2*{sin(-72) + 1 + sin(72) + 1} / 1/2*{cos(-72) + 0} = 2/cos(72) since sin(-72) = -sin(72), and cos(-72) = cos(72) . . . . . (A) Also tan(27) + tan(63) = sin(27)/cos(27) + sin(63)/cos(63) = {sin(27)*cos(63) + sin(63)*cos(27)} / {cos(27)*cos(63)} = 1/2*{sin(-36) + sin(90)} + 1/2*{sin(72) + sin(36)} / 1/2*{cos(-36) + cos(90)} = 1/2*{sin(-36) + 1 + sin(36) + 1} / 1/2*{cos(-36) + 0} = 2/cos(36) since sin(-36) = -sin(36), and cos(-36) = cos(36) . . . . . (B) Therefore, by (A) and (B), tan(9) - tan(27) - tan(63) + tan(81) = tan(9) + tan(81) - tan(27) - tan(63) = 2/cos(72) – 2/cos(36) = 2*{cos(36) – cos(72)} / {cos(72)*cos(36)} = 2*2*sin(54)*sin(18)/{cos(72)*cos(36)} . . . . . . . (C) But cos(72) = sin(90-72) = sin(18) so that sin(18)/cos(72) = 1 and cos(36) = sin(90-36) = sin(54) so that sin(54)/cos(36) = 1 and therefore from C, tan(9) – tan(27) – tan(63) + tan(81) = 2*2*1*1 = 4


What is the height of a building when the distance between its angles of elevation which are 29 degrees and 37 degrees is 30 meters on level ground?

Using trigonometry its height works out as 63 meters to the nearest meter. -------------------------------------------------------------------------------------------------------- let: h = height building α, β be the angles of elevation (29° and 37° in some order) d be the distance between the elevations (30 m). x = distance from building where the elevation of angle α is measured. Then: angle α is an exterior angle to the triangle which contains the position from which angle α is measured, the position from which angle β is measured and the point of the top of the building. Thus angle α = angle β + angle at top of building of this triangle → angle α > angle β as the angle at the top of the building is > 0 → α = 37°, β = 29° Using the tangent trigonometric ratio we can form two equations, one with angle α, one with angle β: tan α = h/x → x = h/tan α tan β = h/(x + d) → x = h/tan β - d → h/tan α = h/tan β - d → h/tan β - 1/tan α = d → h(1/tan β - 1/tan α) = d → h(tan α - tan β)/(tan α tan β) = d → h = (d tan α tan β)/(tan α - tan β) We can now substitute the values of α, β and x in and find the height: h = (30 m × tan 37° × tan 29°)/(tan 37° - tan 29°) ≈ 63 m


The two sides of a right triangle are of length 19 and 63 What is the measure of either of the acute angles in degrees?

Assuming that neither of the given sides is the hypotenuse, then if A is one of the acute angles, tan(A) = 19/63 So A = arctan(19/63) = 16.8 degrees. The other acute angle is 73.2 deg.


What is 50 plus 63 plus 72 plus 63 plus 137 plus 172?

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63 plus 47 plus 42 plus 21 is equal to 173.


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The answer to this problem is 703.


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Angle of elevation tan^-1(14/7) = 63 degrees rounded


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63+59+75 = 197


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12 + 15 + 63 + 12 + 24 + 34 + 23 + 15 = 198


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12 minus 63 plus 9 is -42


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82 + 263 + 17 - 63 = 299


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