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What is 12 b plus 3 equals 39 b equals?
b = 12 12(3)+3 = 39, use a calculator to find out
Can sine theta equals tan theta equals theta be true for small angles?
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.
If a equals b and b equals c then what does c plus b equals?
2a. (a, b and c are all equal.)
If a equals 3 and bequals minus 5 then a minus b equals?
If a equals 3 and b equals minus 5 then a minus b equals what
If A equals 6 what does b equals?
6 === the what does B =
If sin b equals to x over y where b is acute find tan b in terms of x?
tan(b) = x/sqrt(y^2-x^2)
What is tan20tan32 plus tan32tan38 plus tan38tan20?
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
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).
A squared plus b squared equals 26 where a equals b find a and b?
a=b=3.60555
Find the lengths of the missing sides if side a is opposite angle A side b is opposite angle B and side c is the hypotenuse. tan(A) 100 b 100?
Given that ( \tan(A) = \frac{b}{a} ) and ( b = 100 ), we can express ( a ) as ( a = \frac{b}{\tan(A)} = \frac{100}{\tan(A)} ). To find the hypotenuse ( c ), we can use the Pythagorean theorem: ( c = \sqrt{a^2 + b^2} ). Thus, ( c = \sqrt{\left(\frac{100}{\tan(A)}\right)^2 + 100^2} ).
Cos A equals one third. where A is greater than zero but less than pi over two. Sin B equals negative one half. where B is greater than negative pi over two but less than zero. Find Tan A plus B?
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 varies jointly as b and c One set of values is a equals 2.4 b equals 0.6 and c equals 0.8 Find a when b equals 0.4 and c equals 0.4?
a = 0.8
What is the exact trigonometric function value of cot 15 degrees?
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)
What is the measure of angle A to the nearest degree if side a equals 490 and side b equals 960?
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
How do you find the perimeter of a parallelogram with B equals 3.6 cm and S equals 2.2 cm?
It depends on what B is and what S is!
What is 12 b plus 3 equals 39 b equals?
b = 12 12(3)+3 = 39, use a calculator to find out
What is 102a if log2 equals a and log3 equals b?
Find 102a if log2=a and log3=b B has no purpose in this question If a=log2, then 102a =102(log2)
