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the x and y components of vector a are 4m and 6m respectively .the x and y cmponents of vector A+B are 10m and 9m respectively .find the length of B and angle that B makes with the axis.
A=4i+6jA+B=10i+9jB=(A+B)-A=6i+3jlength of B=(36+9)1/2=3(5)1/2let o be angle made by B with X-axis..........Tano= -2 [Tano=-y/x] APPROVE IF USEFUL!!!!!!!!!!!!!!!
how is it possible.....here angle makes with x axis but not both x and y axis.... how we use tan(theta) here....i think formula is given cos(theta) or sin(theta)depends on coordinate system
A=4i+6jA+B=10i+9jB=(A+B)-A=6i+3jlength of B=(36+9)1/2=3(5)1/2let o be angle made by B with X-axisthus O=tan^-1(-1/2) [Tano=-y/x]
but, how can it be negative ...i mean tan^-1(-1/2) , also in my book ans is tan^-1(1/2) can u explain?
The above ans for the length of B is correct (45^1/2 or 3(5)^1/2)... But for tan(theta) it would be.. Tan(theta) = By/Bx = 3/6 = 1/2... Therefore, theta = tan^-1(1/2)
Hello StudentPlease see the solution in the attachment.I hope this answer will help you.
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