Trigonometry (SOHCAHTOA): Question 1

Worked solution

Setting up the triangle

Sketch the right-angled triangle $PFT$ with the right angle at the foot $F$:

• $PF = 28\text{ m}$ is the horizontal distance; this is the side adjacent to the $37^\circ$ angle at $P$.
• $TF$ is the vertical height; this is the side opposite the $37^\circ$ angle.
• $PT$ is the sloping line of sight; this is the hypotenuse.

Recall SOH-CAH-TOA:

$\sin\theta=\frac{\text{opp}}{\text{hyp}},\qquad \cos\theta=\frac{\text{adj}}{\text{hyp}},\qquad \tan\theta=\frac{\text{opp}}{\text{adj}}.$

---

Part (a): the height $TF$

We know the adjacent side ($28\text{ m}$) and we want the opposite side ($TF$). The ratio linking opposite and adjacent is the tangent:

$\tan 37^\circ=\frac{\text{opp}}{\text{adj}}=\frac{TF}{28}.$

Multiply both sides by $28$:

$TF = 28\,\tan 37^\circ.$

Make sure the calculator is in degree mode. Since $\tan 37^\circ = 0.75355\ldots$,

$TF = 28 \times 0.75355\ldots = 21.0995\ldots\text{ m}.$

Rounding to 1 decimal place:

$\boxed{TF \approx 21.1\text{ m}}$

---

Part (b): the distance $PT$

Now we want the hypotenuse $PT$. We can use the adjacent side $28\text{ m}$ with the cosine ratio:

$\cos 37^\circ=\frac{\text{adj}}{\text{hyp}}=\frac{28}{PT}.$

Rearrange to make $PT$ the subject. Multiply both sides by $PT$, then divide by $\cos 37^\circ$:

$PT=\frac{28}{\cos 37^\circ}=\frac{28}{0.79863\ldots}=35.06\ldots\text{ m}.$

Rounding to the nearest metre:

$\boxed{PT \approx 35\text{ m}}$

Alternative method (using part (a) and Pythagoras)

Once you have the height $TF \approx 21.1\text{ m}$, you can find the hypotenuse with Pythagoras’ theorem:

$PT=\sqrt{PF^{2}+TF^{2}}=\sqrt{28^{2}+21.0995^{2}}=\sqrt{784+445.19}=\sqrt{1229.19}=35.06\ldots\text{ m},$

which again gives $PT \approx 35\text{ m}$. (Using the unrounded value of $TF$ avoids rounding error.) A third option is $PT=\dfrac{TF}{\sin 37^\circ}=\dfrac{21.0995}{0.60182}\approx 35\text{ m}$; all three routes agree.

---

Final answers

• (a) $TF \approx 21.1\text{ m}$
• (b) $PT \approx 35\text{ m}$