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What is direction of force?
b
What does the direction of arrow b indicate?
knee
If one component of a vector A is zero along the direction of another vector B then in what direction the two vectors will be?
opposite direction.
What kind of movement is the arm engaged in when going from position A to position B?
extention
How do you solve for the resultant if two forces are in the same direction?
If two forces are in the same direction, then their resultant is also in the same direction, and its magnitude is the sum of the two components' magnitudes.
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).
How do you find tan B if tan A equals 678?
tan A says nothing about tan B without further information.
How to find the direction of vector A by its x and y components?
Consider any two points on the vector, P = (a, b) and Q = (c, d). And lext x be the angle made by the vector with the positive direction of the x-axis. Then either a = c, so that the vector is vertical and its direction is straight up or a - c is non-zero. In that case, tan(x) = (b - d)/(a - c) or x = tan-1[(b - d)/(a - c)]
Why magnetism at the centre of a bar magnet is zero?
F = mB - mB =0 a bar magnet is placed in a uniform magnetic field B, its poles +m and -m experience force mB and mB along and opposite to the direction of magnetic field B.
The function cot 115 degrees is equivalent to A -tan 65 degrees B -tan 25 degrees C tan 25 degrees D tan 65 degrees?
B: -tan(25)
Diploma First Year Maths Important Questions List?
tan a=1/2 tan b=1/3 find tan (a-b)
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)
How do you find the poles of a magnet?
If there is a repulsion between A and N then A is North pole and B is South pole of the horse shoe magnet. If B and N repel each other the B is north and A is south of the horse shoe magnet.
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)
How do you find the poles of a horseshoe magnet?
If there is a repulsion between A and N then A is North pole and B is South pole of the horse shoe magnet. If B and N repel each other the B is north and A is south of the horse shoe magnet.
What is the word that describes the distance between your final position and your starting position?
That's the magnitude of 'Displacement'.If you want the complete 'Displacement', you also have to includethe direction from Point-A to Point-B.
