Find the vertical and horizontal component ? Steps please

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2 Answers
Aug 30, 2017

V_H = 8.62 and V_V = 11.03 to 2 dp

Explanation:

This is just trigonometry. We know that

sin(theta) = "opposite"/"hypotenuse" and cos(theta) = "adjacent"/"hypotenuse"

So for all problems like this we can instantly see that the opposite corresponds to the vertical and the adjacent corresponds to the horizontal.

V_H = abs(vec V)cos(theta) and V_V = abs(vec V) sin(theta)

In this case, the magnitude is 14 and theta = 52.0

V_H = 14*cos(52) = 8.62 and V_V = 14*sin(52) = 11.03

Aug 30, 2017

V_v approx 11.032 (3 decimal places)
V_h approx 8.619 (3 decimal places)

Explanation:

You may recall SOHCAHTOA for a right angled triangle from trigonometry:

H = Hypotenuse, O = Opposite side, A = Adjacent side

sin(theta) = (O)/(H), cos(theta) = A/H, tan(theta) = (O)/(A)

In this case, the horizontal component V_h is adjacent to the angle, and the vertical component V_v is opposite the angle. In both cases, theta is given as 52° and the hypotenuse is given as 14.0.enter image source here

We can note the two relationships to solve the vector into components:

sin(52) = (O)/H = (V_v)/14

cos(52) = A/H = (V_h)/14

rearranging for V_v and V_h,

V_v = sin(52)*14 approx 11.03215055 approx 11.032 (3 decimal places)
V_h = cos(52)*14 approx 8.619260655 approx 8.619 (3 decimal places)