b = 12 12(3)+3 = 39, use a calculator to find out
2a. (a, b and c are all equal.)
If a equals 3 and b equals minus 5 then a minus b equals what
If sin θ = tan θ, that means cos θ is 1 (since tan θ = (sin θ)/(cos θ)) (Usually in and equation a/b=a, b doesn't have to be 1 when a is 0, but cos θ = 1 if and only if sin θ = 0) The angles that satisfy cos θ = 1 is 2n(pi) (or 360n in degrees) When n is an integer. But if sin θ = tan θ = θ, the only answer is θ = 0. Because sin 0 is 0 and cos 0 is 1 and tan 0 is 0 The only answer would be when θ = 0.
6 === the what does B =
tan(b) = x/sqrt(y^2-x^2)
Oh honey, you're throwing some trigonometry at me? Alright, buckle up. The sum of tan20tan32 plus tan32tan38 plus tan38tan20 is equal to 1. Just plug in those values and watch the magic happen. Math can be sassy too, you know.
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).
tan a=1/2 tan b=1/3 find tan (a-b)
a=b=3.60555
Ok, im not positive about this but here is what i got. (ca=cos(a),sa=sin(a),cb=cos(b)ect) ca=1/3 | ca^2=1/9 | sa^2=1-ca^2 | sa^2=8/9 | sa=.943 Sin(a)=.943 Cos(a)=.333 Tan(a)=2.832 sb=-1/2 | sb^2=1/4 | cb^2=1-.25 | cb^2=3/4 | cb=-.866 Sin(b)=-.5 Cos(b)=-.866 Tan(b)=.577 Tan(a+b)=(Tan(a)+Tan(b))/(1-2.832*.577) Tan(a+b)=3.41/-.634 Tan(a+b)=-5.376 So that's what i got. If you have any questions or if you got a different answer you should email me. actually email me regardless, i want to see if i got it right. allonbacuth@gmail.com
a = 0.8
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)
It depends on what B is and what S is!
b = 12 12(3)+3 = 39, use a calculator to find out
There is not enough information. Angle A can be anything between 0 and 180 degrees. If you are talking about a triangle ABC, in which angle C is the right angle then a is the opposite side and b is the adjacent side. tan A = opposite / adjacent tan A = 490 / 960 = 0.51041666666666666666666666666667 and A = invtan(0.51041666666666666666666666666667) [also called tan-1] so A = 27.04 degrees
Find 102a if log2=a and log3=b B has no purpose in this question If a=log2, then 102a =102(log2)