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tan A says nothing about tan B without further information.

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โˆ™ 2010-12-10 16:42:38
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Q: How do you find tan B if tan A equals 678?
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If sin b equals to x over y where b is acute find tan b in terms of x?

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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).

Diploma First Year Maths Important Questions List?

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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

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?

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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.

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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)

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What is the measure of angle A to the nearest degree if side a equals 490 and side b equals 960?

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Unfortunately, limitations of the browser used by means that we cannot see most symbols. It is therefore impossible to give a proper answer to your question. Please resubmit your question spelling out the symbols as "plus", "minus", "times", "equals".

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