Yes. There are 100 centimeters in a meter. centi means 1 hundred. For example century is 100 years.
Use the tangent ratio: 23*tan(23) = 9.762920773 Answer: 10 meters to the nearest meter
tan(9) + tan(81) - tan(27) - tan(63) = 4
Tan Tan
This may not be the most efficient method but ... Let the three angle be A, B and C. Then note that A + B + C = 20+32+38 = 90 so that C = 90-A+B. Therefore, sin(C) = sin[(90-(A+B) = cos(A+B) and cos(C) = cos[(90-(A+B) = sin(A+B). So that tan(C) = sin(C)/cos(C) = cos(A+B) / sin(A+B) = cot(A+B) Now, tan(A+B) = [tan(A)+tan(B)] / [1- tan(A)*tan(B)] so cot(A+B) = [1- tan(A)*tan(B)] / [tan(A)+tan(B)] The given expressin is tan(A)*tan(B) + tan(B)*tan(C) + tan(C)*tan(A) = tan(A)*tan(B) + [tan(B) + tan(A)]*cot(A+B) substituting for cot(A+B) gives = tan(A)*tan(B) + [tan(B) + tan(A)]*[1- tan(A)*tan(B)]/[tan(A)+tan(B)] cancelling [tan(B) + tan(A)] and [tan(A) + tan(B)], which are equal, in the second expression. = tan(A)*tan(B) + [1- tan(A)*tan(B)] = 1
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).
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 Cerca...Tan Lejos was created in 1975.
The airport code for Tan Tan Airport is TTA.
cot(15)=1/tan(15) Let us find tan(15) tan(15)=tan(45-30) tan(a-b) = (tan(a)-tan(b))/(1+tan(a)tan(b)) tan(45-30)= (tan(45)-tan(30))/(1+tan(45)tan(30)) substitute tan(45)=1 and tan(30)=1/√3 into the equation. tan(45-30) = (1- 1/√3) / (1+1/√3) =(√3-1)/(√3+1) The exact value of cot(15) is the reciprocal of the above which is: (√3+1) /(√3-1)
If the angles are measured in degrees or gradians, then: tan 3 > tan 2 > tan 1 If the angles are measured in radians, then: tan 1 > tan 3 > tan 2.
tan(135) = -tan(180-135) = -tan(45) = -1