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AP Physics 1: Algebra Notes

1.5.3 Using Trigonometry to Resolve Vectors

AP Syllabus focus: ‘Perpendicular vector components can be found using trigonometric relationships and the Pythagorean theorem.’

Vectors in two dimensions are often easier to use when split into perpendicular parts. This page focuses on using right-triangle trigonometry and the Pythagorean theorem to move between a vector’s magnitude/angle and its components.

Why resolve vectors into components?

Many physics quantities are vectors (force, velocity, displacement). In 2D, you can treat one vector as two independent, perpendicular “shadow” vectors along chosen axes (often xx and yy). This lets you apply algebra separately in each direction.

Components (what you are finding)

Components: The perpendicular vector parts (e.g., AxA_x and AyA_y) whose vector sum equals the original vector A\vec{A}.

Components are not extra vectors added to the situation; they are just an equivalent representation of the same vector.

The right-triangle picture

Resolving a vector uses a right triangle:

Pasted image

A 2D vector is drawn as the hypotenuse of a right triangle, with the legs representing the perpendicular components along the coordinate axes. The labels show how the components relate to the magnitude via Ax=AcosθA_x = A\cos\theta and Ay=AsinθA_y = A\sin\theta when θ\theta is measured from the +xx axis. This picture helps prevent the common sine/cosine swap by making “adjacent” and “opposite” explicit. Source

  • The hypotenuse is the vector’s magnitude AA.

  • The legs are the component magnitudes Ax|A_x| and Ay|A_y|.

  • The angle θ\theta is measured from one axis to the vector (you must state which axis you’re using).

Trig relationships for components

If θ\theta is measured from the +xx axis toward the vector, then:

  • The adjacent side corresponds to the xx-component.

  • The opposite side corresponds to the yy-component.

Use cosine for adjacent/hypotenuse and sine for opposite/hypotenuse. The sign of each component comes from direction (left/down negative if your axes define them that way), not from the trig function alone.

Resolve components: Ax=Acosθ,Ay=Asinθ \text{Resolve components: } A_x = A\cos\theta,\quad A_y = A\sin\theta

AA = magnitude of the vector (units of the vector)

AxA_x = xx-component (same units as AA)

AyA_y = yy-component (same units as AA)

θ\theta = angle from +xx axis to the vector (radians or degrees, consistent with your calculator)

Recombine (Pythagorean): A=Ax2+Ay2 \text{Recombine (Pythagorean): } A = \sqrt{A_x^2 + A_y^2}

AA = magnitude of the vector (units of the vector)

AxA_x = xx-component (same units as AA)

AyA_y = yy-component (same units as AA)

A common mistake is swapping sine and cosine because the angle is actually measured from the yy axis. If θ\theta is from +yy, then the “adjacent” side is Ay|A_y|, so the roles of sine/cosine switch.

Using the Pythagorean theorem appropriately

The Pythagorean theorem applies only when the components are perpendicular:

  • Valid: xx and yy axes (90° apart)

  • Not valid: two components chosen along non-perpendicular directions

When you compute A=Ax2+Ay2A=\sqrt{A_x^2+A_y^2}, the squared terms remove signs; magnitude is always non-negative.

Pasted image

The vector and its perpendicular components are shown forming a right triangle, so the magnitude follows from the Pythagorean theorem: A=Ax2+Ay2A = \sqrt{A_x^2 + A_y^2}. By tying the geometry to the algebra, the figure clarifies why this magnitude formula is valid specifically for orthogonal xxyy components. It also reinforces that component signs affect direction, not the nonnegative magnitude. Source

Direction must be handled separately.

Angle from components (direction)

To find direction from known components, the basic trig idea is:

  • tanθ=AyAx\tan\theta = \dfrac{A_y}{A_x} (when θ\theta is from +xx)

However, the correct direction depends on the quadrant (signs of AxA_x and AyA_y). Many calculators provide an “atan2” function (or equivalent) to determine the correct quadrant; otherwise, you must adjust the angle based on signs.

Checklist for AP Physics 1 Algebra problems

  • Choose axes and define + directions clearly.

  • Identify the angle’s reference axis (from +xx or +yy).

  • Use sine/cosine based on opposite/adjacent relative to that angle.

  • Assign signs to AxA_x and AyA_y from direction.

  • Use the Pythagorean theorem only for perpendicular components.

FAQ

Identify which component is adjacent to the given angle.

  • If the angle is from +$y$, then $|A_y|=A\cos\theta$ and $|A_x|=A\sin\theta$.

  • Then apply signs based on left/right and up/down.

Switch the calculator mode to match the problem or convert units.

  • Degrees to radians: multiply by $\pi/180$.

  • Radians to degrees: multiply by $180/\pi$. Using the wrong mode typically gives a plausible-looking but incorrect component.

Angles are periodic: adding $360^\circ$ (or $2\pi$) gives the same direction. Also, a negative angle (clockwise) can represent the same direction as a positive anticlockwise angle, e.g., $-45^\circ$ equals $315^\circ$.

Check signs first: $(A_x,A_y)$ determines the quadrant. Then ensure the final angle places the vector in that quadrant. If available, use $\text{atan2}(A_y,A_x)$, which returns an angle consistent with both signs.

Yes for perpendicular components: $|A_x|\le A$ and $|A_y|\le A$ because $| \cos\theta |\le 1$ and $| \sin\theta |\le 1$. If a computed component exceeds $A$, it signals a trig/angle/reference-axis error.

Practice Questions

Q1 (1–3 marks) A vector F\vec{F} has magnitude 20 N20\ \text{N} at an angle of 3030^\circ above the +xx axis. State FxF_x and FyF_y.

  • Fx=20cos30F_x = 20\cos30^\circ (1)

  • Fy=20sin30F_y = 20\sin30^\circ (1)

  • Correct values with units: Fx17.3 NF_x \approx 17.3\ \text{N}, Fy=10.0 NF_y = 10.0\ \text{N} (1)

Q2 (4–6 marks) A displacement has components sx=6.0 ms_x = -6.0\ \text{m} and sy=8.0 ms_y = 8.0\ \text{m}. Find (i) the magnitude ss and (ii) the direction as an angle measured anticlockwise from the +xx axis.

  • Use s=sx2+sy2s=\sqrt{s_x^2+s_y^2} (1)

  • Substitute: s=(6.0)2+8.02s=\sqrt{(-6.0)^2+8.0^2} (1)

  • Correct magnitude: s=10.0 ms=10.0\ \text{m} (1)

  • Use tanθ=sy/sx\tan\theta = s_y/s_x or θ=tan1(sy/sx)\theta=\tan^{-1}(s_y/s_x) (1)

  • Recognise quadrant (negative xx, positive yy) and adjust angle (1)

  • Correct direction: θ126.9\theta \approx 126.9^\circ from +xx (1)

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