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In a right triangle, where the angle "x" is adjacent to the hypotenuse, the tangent of that angle would be the length of the opposite side divided by the length of the adjacent side (not the hypotenuse).
For instance, consider a right triangle with the following side lengths:
A = 3
B = 4
C = 5
B is the hypotenuse. The tangent of the angle between sides A and C would be B/A, or 4/3. The tangent of the angle between sides B and C would be B/A, or 3/4.
This is often taught with the memorisation acronym "SOHCAHTOA":
(S)ine = (O)pposite / (H)ypotenuse
(C)osine = (A)djacent / (H)ypotenuse
(T)angent = (O)pposite / (A)djacent
What else can I help you with?
What is Tan formula?
The tangent formula for a right angle triangle is tangent = opposite/adjacent20^\circ20​∘​​77??CCBBAA
What is cos x tan x simlpified?
The definition of tan(x) = sin(x)/cos(x). By this property, cos(x)tan(x) = sin(x).
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).
Verify the identity sinx cotx - cosx divided by tanx equals 0?
(sin(x)cot(x) - cos(x))/tan(x)(Multiply by tan(x)/tan(x))sin(x) - cos(x)tan(x)(tan(x) = sin(x)/cos(x))sinx - cos(x)(sin(x)/cos(x))(cos(x) cancels out)sin(x) - sin(x)0
What is dydx of sincostanx?
If y = sin(cos(tan(x))) Using the chain rule: (f(g(x)))' = f'(g(x)).g'(x) Then dy/dx = cos(cos(tan(x))).-sin(tan(x)).sec2(x) = -cos(cos(tan(x))).sin(tan(x)).sec2(x) Unfortunately I don't think this can be simplified much more. ( sec = 1/cos )
