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Wize University Physics Textbook (Master) > Force Vectors (Advanced Info)

Basic Trigonometric Rules - Review

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For RIGHT triangles: “ SOH CAH TOA ”

(i like to remember SINE as "SIDE OPPOSITE", and COSINE as "the CLOSE" side)



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For NON-RIGHT triangles:


What about that small TRIANGLE with #'s?


On occasion you will see vectors presented as above. Students often miss the EASY way to handle this when finding x & y components of the 2D vector. They will want to find the angle using SOHCAHTOA, which is fine, but is "one more calculation and one more place to possibly make an error" that would carry through the rest of the solution. My suggestion, use this trick: using the example above, cos(theta) = adjacent/hypotenuse = 4/5. Instead of solving for the angle theta (which you will only reuse in cos(theta) or sin(theta)), why not just use the "4/5" instead of cos(theta). For example, if the vector above was 10N, then:

Fx = -10 cos(theta) N

Instead, just write as:

Fx = -10(4/5) = -8N And then Fy = 10(3/5) = 6N

And last, but not least.......... make sure your calculator is in DEGREES, and not radians!
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Example: Solving for Missing Angles

Find the missing angles in this triangle.

Let's start by finding x\bco x

The give sides are the opposite side and hypotenuse side, so we use the sin⁡\sin ratio:
sin⁡x=OppHypsin⁡x=715x=sin⁡−1(715)x≈27.82°\begin{array}{rcl} \sin x&=&\dfrac{Opp}{Hyp}\\[1em] \sin x&=&\dfrac{7}{15}\\[1em] x&=&\sin^{-1}\left(\dfrac{7}{15}\right)\\[1em] x &\approx&27.82\degree \end{array}

Finding y\bco y

Method 1 (easier way)
The interior angles in a triangle must add up to 180°180\degree:
x+y+90°=180°27.82+y+90°=180°y+117.82°=180°y=180°−117.82°y=62.18°\begin{array}{rcl} x+y+90\degree&=&180\degree\\ 27.82+y+90\degree&=&180\degree\\ y+117.82\degree&=&180\degree\\ y&=&180\degree-117.82\degree\\ y&=&62.18\degree \end{array}

Method 2 (using inverse trig ratios again)
We could use the cos⁡\cos ratio:
cos⁡y=AdjHypcos⁡y=715y=cos⁡−1(715)y≈62.18°\begin{array}{rcl} \cos y&=&\dfrac{Adj}{Hyp}\\[1em] \cos y&=&\dfrac{7}{15}\\[1em] y&=&\cos^{-1}\left(\dfrac{7}{15}\right)\\[1em] y &\approx&62.18\degree \end{array}
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Example: Finding a Side Length in a Right Triangle

Find the missing side length xx in each of the following triangle:

a)

Since we are given the Hypotenuse and we want to find the Opposite, we must use the sin⁡\sin ratio.
sin⁡θ=OppHypsin⁡40°=x5sin⁡40°×5=x5×5sin⁡40°×5=x0.642×5≈x3.21≈xx≈3.21\begin{array}{rcl} \sin\theta&=&\dfrac{Opp}{Hyp}\\[1em] \sin40\degree&=&\dfrac{x}{5}\\[1em] \sin40\degree\colorTwo{\times5}&=&\dfrac{x}{5}\colorTwo{\times5}\\[1em] \sin40\degree\times5&=&x\\[1em] 0.642\times5&\approx&x\\[1em] 3.21&\approx&x\\[1em] x&\approx&3.21 \end{array}

Therefore, the exact value of xx is 5×sin⁡40°5\times \sin40\degree, which is approximately 3.21.
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b)

Since we are given the Adjacent and we want to find the Hypotenuse, we must use the cos⁡\cos ratio.
cos⁡θ=AdjHypcos⁡35°=4xcos⁡35°×x=4x×xcos⁡35°×x=4cos⁡35°×xcos⁡35°=4cos⁡35°x≈40.819x≈4.88\begin{array}{rcl} \cos\theta&=&\dfrac{Adj}{Hyp}\\[1em] \cos35\degree&=&\dfrac{4}{x}\\[1em] \cos35\degree\colorTwo{\times x}&=&\dfrac{4}{x}\colorTwo{\times x}\\[1em] \cos35\degree\times x&=&4\\[1em] \dfrac{\cos35\degree\times x}{\colorTwo{\cos35\degree}}&=&\dfrac{4}{\colorTwo{\cos35\degree}}\\[1em] x&\approx&\dfrac{4}{0.819}\\[1em] x&\approx&4.88\\ \end{array}

Therefore, the exact value of xx is 4cos⁡35°\dfrac{4}{\cos35\degree}, which is approximately 4.88.

Practice: Labelling a Right-Angle Triangle

Label the sides of the following triangle with O (for opposite), A (for adjacent), and H (for hypotenuse) according to the given angle.


Practice: Primary Trig Ratios

Given the following triangle, find all of the primary trig ratios.

Find the length of BD and angle C.

BD: since we have 2 of the 3 sides of the right triangle, we can use Pythagorean's Theorem to solve for BC.
BD2 + 32 = 52
BD = sqrt (25 - 9) = 4 ANS. (note this 3-4-5 right triangle is sometimes called a Pythagorean Triple)

Angle C: using trig, we can find angle C. we can use cosine, sine, or tangent to solve. HOWEVER, is we use the given sides of 3 and 5, we don't risk using the side BD which was calculated and not given. ie. if we made a mistake in finding BD, the angle would be incorrect too.
cos(C) = 3/5
angle C = cos-1(3/5) = 53° ANS.


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Find the length of AB and angle B (formed by segments AB and BD).