Trigonometric Ratios

Suppose we have the below triangle, find the side labelled x.

Pythagoras- Example Finding Hypotenuse, Jordan Madge- StudySmarter Originals

Solution:

First, let's label each of the sides a, b and c:

Now, Pythagoras' theorem states that $a^2+b^2=c^2$.

Substituting $a=3,b=4,c=x$ into this, we obtain $9+16=x^2$, or$x^2=25$.

Therefore, to find x we simply take the square root both sides. Thus, $x=5cm$.

Suppose we have the below triangle, find the side labelled x.

Solution:

First, let's label each of the sides a, b and c:

Now, this example is slightly different from the previous as this time we are not finding the hypotenuse, c. Therefore, we have to use a rearranged version of Pythagoras' theorem: $b^2=c^2-a^2$. Substituting $a=2,b=x$ and $c=5$ into this formula, we obtain $x^2=5^2-2^2=25-4=21$. Thus, we simply need to square root both sides to obtain that $x=\sqrt{21}=4.58cm$ (2.d.p).

Suppose we have the below triangle. Find the side labelled x.

Trigonometry- Example Finding Missing Side, Jordan Madge- StudySmarter Originals

Solution:

First, let's label each of the sides O, A and H.

The next step is to work out which two of the three sides are involved. What we mean by this, is which sides do we know or wish to know. In this example, we know the hypotenuse is $3cm$ and we want to know the adjacent side, $xcm$. Thus, the two sides involved are the adjacent and hypotenuse. Next, we must identify which trigonometric Ratio involves the adjacent and hypotenuse. In this case, it is cos, because cos is the only identity that involved both the adjacent and hypotenuse side. Substituting $\theta =42,A=x,H=3$ into $\cos \left(\theta \right)=\frac{A}{H}$we obtain $\cos \left(42\right)=\frac{x}{3}$.

All we have to do now is a bit of rearranging to find x. We can do this by multiplying both sides by 3. Thus,$x=3\cos \left(42\right)$. We get the final answer by typing this into our calculator. So, $x=2.23cm.$

Suppose we have the below triangle. Find the side labelled x.

Trigonometry- Example Finding Missing Side, Jordan Madge- StudySmarter Originals

Solution:

First, let's label each of the sides O, A and H.

In this case, we know the hypotenuse, and we wish to know the opposite side. Therefore we use sin since sine is the only trigonometric ratio involving the hypotenuse and opposite side.

Suppose we have the below triangle. Find the side labelled x.

Trigonometry- Example Finding Missing Side, Jordan Madge- StudySmarter Originals

Solution:

First, let's label each of the sides O, A and H.

In this case, we know the adjacent side, and we wish to know the hypotenuse. Therefore we use cos.

Since $\cos \left(\theta \right)=\frac{A}{H}$ we have $\cos \left(51\right)=\frac{8}{x}$.

Now, to rearrange this, we need to first multiply both sides by x to obtain $x\cos \left(51\right)=8$. Then, to find x, we divide both sides by $\cos \left(51\right)$ to get $x=\frac{8}{\cos \left(51\right)}=12.7cm$(3.s.f).

Suppose we have the below triangle. Find the side labelled x.

Trigonometry- Example Finding Missing Side, Jordan Madge- StudySmarter Originals

Solution:

First, let's label each of the sides O, A and H.

In this case, we know the opposite side, and we wish to know the adjacent side. Therefore we use tan.

Since $\tan \left(\theta \right)=\frac{O}{A}$ we have that $\tan \left(30\right)=\frac{3}{x}$. Multiplying both sides by x, we have $x\tan \left(30\right)=3$. Dividing both sides by$\tan \left(30\right)$ we get $x=\frac{3}{\tan \left(30\right)}=5.20cm$ (3.s.f).

Using your calculator, find ${\sin }^{-1}(0.1),{\cos }^{-1}(0.1)$ and ${\tan }^{-1}(0.1)$.

Solution:

${\sin }^{-1}(0.1)=shift+\sin (0.1)=5.{74}^{°}$(3.s.f)

${\cos }^{-1}(0.1)=shift+\cos (0.1)=84.3^{°}$ (3.s.f)

${\tan }^{-1}(0.1)=shift+\tan (0.1)=5.{71}^{°}$ (3.s.f)

Suppose we have the below triangle. Find the angle labelled $\theta$.

Trigonometry- Example Finding Missing Angle, Jordan Madge- StudySmarter Originals

Solution:

To find missing angles, the steps are pretty much the same as before. However, there is one minor difference. As before, let's start by labelling each of the sides O, A and H.

Now, we again need to identify which sides are involved. In this case, we know the adjacent and hypotenuse. Since cos involves adjacent and hypotenuse, we use the cosine.

Since$\cos \left(\theta \right)=\frac{A}{H}$, we have $\cos \left(\theta \right)=\frac{1.2}{3}$.

Now this time, to get theta by itself, we must take the inverse cosine of both sides.

Therefore, our answer is $\theta ={\cos }^{-1}\left(\frac{1.2}{3}\right)=66.4^{°}$ (3.s.f).

Suppose we have the below triangle. Find the angle labelled $\theta$.

Trigonometry- Example Finding Missing Angle, Jordan Madge- StudySmarter Originals

Solution:

Labelling the sides, we can see that we have the opposite side and hypotenuse. Therefore we use sine.

Since$\sin \left(\theta \right)=\frac{O}{H}$, we have $\sin \left(\theta \right)=\frac{2.8}{5.2}$

To get $\theta$ by itself, we must take the inverse sine of both sides. Thus, $\theta ={\sin }^{-1}\left(\frac{2.8}{5.2}\right)=32.6^{°}$ (3.s.f).

Suppose we have the below triangle. Find the side labelled x.

(non-calculator question)

Trigonometry- Example Finding Missing Side, Jordan Madge- StudySmarter Originals

Solution:

Since we want to know the opposite side and we have the opposite side, we will use sine.

Since $\sin \left(\theta \right)=\frac{O}{H}$, we have $\sin \left(30\right)=\frac{x}{2}$.

Rearranging, we find that $x=2\sin \left(30\right)$.

Using table 1, we see that $\sin \left(30\right)=\frac{1}{2}$

Thus, $x=2\times \frac{1}{2}=1cm.$

Suppose we have the below triangle. Find the angle labelled $\theta$.

(non-calculator question)

Solution:

Since we have been given both the opposite and adjacent sides, we use tan.

Since $\tan \left(\theta \right)=\frac{O}{A}$, we have that $\tan \left(\theta \right)=\frac{6.3}{6.3}=1$.

To get $\theta$ by itself, we must take the inverse tan of both sides.

Thus, $\theta ={\tan }^{-1}\left(1\right)$

Using table 1, we see that $\tan \left(45\right)=1$and thus $\theta ={45}^{°}$.