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What is the value of tan 1?
tan(1) = 1.5574 where the angle is measured in radians.
What is the value of tan 39?
tan(39c) = 3.6146 If the angle was not measured in radians (the mathematical standard), you should have mentioned the units used.
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
Tan 9 plus tan 81 -tan 27-tan 63?
tan(9) + tan(81) - tan(27) - tan(63) = 4
If for a triangle abc tan a-b plus tan b-c plus tan c-a equals 0 then what can you say about the triangle?
tan (A-B) + tan (B-C) + tan (C-A)=0 tan (A-B) + tan (B-C) - tan (A-C)=0 tan (A-B) + tan (B-C) = tan (A-C) (A-B) + (B-C) = A-C So we can solve tan (A-B) + tan (B-C) = tan (A-C) by first solving tan x + tan y = tan (x+y) and then substituting x = A-B and y = B-C. tan (x+y) = (tan x + tan y)/(1 - tan x tan y) So tan x + tan y = (tan x + tan y)/(1 - tan x tan y) (tan x + tan y)tan x tan y = 0 So, tan x = 0 or tan y = 0 or tan x = - tan y tan(A-B) = 0 or tan(B-C) = 0 or tan(A-B) = - tan(B-C) tan(A-B) = 0 or tan(B-C) = 0 or tan(A-B) = tan(C-B) A, B and C are all angles of a triangle, so are all in the range (0, pi). So A-B and B-C are in the range (- pi, pi). At this point I sketched a graph of y = tan x (- pi < x < pi) By inspection I can see that: A-B = 0 or B-C = 0 or A-B = C-B or A-B = C-B +/- pi A = B or B = C or A = C or A = C +/- pi But A and C are both in the range (0, pi) so A = C +/- pi has no solution So A = B or B = C or A = C A triangle ABC has the property that tan (A-B) + tan (B-C) + tan (C-A)=0 if and only if it is isosceles (or equilateral).
