Trigonometry Question (finding the sin, cos, cosec etc on a right-angled triangle)

Question

For the right-angled triangle $\widehat{PQR}$, where $\overline{PQ} = 9\text{ cm}$, $\overline{QR} = 40\text{ cm}$ and $\overline{PR} = 41\text{ cm}$, give the value of:

a) $\sin \hat{P}$ b) $\cos \hat{R}$ c) $\operatorname{cosec} \hat{P}$ d) $\cot \hat{R}$ e) $\angle \hat{QRP}$

Answer 1

In case triangle it is useful to draw our right triangle:

Now let us remember the trig definitions for any given angle $\theta$: $$\begin{align} \sin(\theta) &= \frac{\mathrm{opposite}}{\mathrm{hypotenuse}}\ \ &\csc(\theta) &= \frac{\mathrm{hypotenuse}}{\mathrm{opposite}}\\ \cos(\theta) &= \frac{\mathrm{adjacent}}{\mathrm{hypotenuse}} &\sec(\theta) &= \frac{\mathrm{hypotenuse}}{\mathrm{adjacent}}\\ \tan(\theta) &= \frac{\mathrm{opposite}}{\mathrm{adjacent}} &\cot(\theta) &= \frac{\mathrm{adjacent}}{\mathrm{opposite}}\\ \end{align} $$

And so now to solve your problems

Find the $\sin(\beta)$

We know from above that $\sin(\beta) = \frac{\mathrm{opposite}}{\mathrm{hypotenuse}}$, and so we see that side $\overline{QR}$ is opposite to $\beta$ and our hypotenuse is $\overline{RP}$, so we see that $$\sin(\beta) = \frac{40}{41}\ \ .$$

Find the $\cos(\delta)$

We know from above that $\cos(\delta) = \frac{\mathrm{adjacent}}{\mathrm{hypotenuse}}$, and we observe that $\overline{RQ}$ is adjacent to $\delta$, and ur hypotenuse is $\overline{RP}$, so we see that $$ \cos(\delta) = \frac{40}{41}\ \ .†$$

Find the $\csc(\beta)$

We know from above that $csc(\beta) = \frac{\mathrm{hypotenuse}}{\mathrm{opposite}}$, and like observed before, $\overline{QR}$ is opposite to $\beta$ and our hypotenuse is $\overline{RP}$, so we see that $$\csc(\beta) = \frac{41}{40}\ \ .$$

Find the $\cot(\delta)$

We know from above that $\cot(\theta) = \frac{\mathrm{adjacent}}{\mathrm{opposite}}$, and we know that $\overline{RQ}$ is adjacent to $\delta$ and $\overline{QP}$ is opposite $\delta$ and so $$\cot(\theta) = \frac{40}{9}\ \ .$$

Find the angle measure of $\delta $

Finally we have to find the angle measure $\delta$. First let me introduce inverse functions:

Let's say we have some function $f(x) = y$. If we want find what our $x$ is, and we know what $y$ is, we have to take the inverse function written as $f^{-1}(x)$. We define this function as the function that has the property $f^{-1}(f(x)) = x$. So for trig functions, if we know (for example) $\sin(\theta) = y$ then $\sin^{-1}(y) = \theta$.†† You may have seen the $\sin^{-1}$, $\cos^{-1}$, and $\tan^{-1}$ buttons on your calculator, and that is what they are. Just to be fully clear $\sin^{-1}(x) \not = \frac{1}{\sin(x)}$

Now on to the problem, we know what the $\cos(\delta) = \frac{40}{41}$, and so to find $\delta$ we must take the inverses of both sides: $$ \begin{align} \cos^{-1}(\cos(\delta)) &= \cos^{-1}(\frac{40}{41})\\ \delta &= \cos^{-1}(\frac{40}{41})\\ &\approx 12.68^\circ \end{align} $$

That last result is simply evaluated with a calculator.

---

†It is interesting to note that the $\sin(\beta) = \cos(\delta)$, and if we note that $\delta = 90^{\circ} - \beta$ we get the general identity $\sin(\beta) = \cos(90^{\circ} - \beta)$ for any given angle $\beta$

†† The inverse sine (i.e. $\sin^{-1}(\theta)$) is sometimes written as $\arcsin(\theta)$.

Answer 2

$40^2 + 9^2 = 1600 + 81 = 1681$ And $41^2 = 1681$

So this is a right angled triangle (Pythagoras Theorem) Hypot =$ 41$ Base = $9$ Perpendicular = $40$

$\sin = \frac{P}{H}$

$\cos = \frac{B}{H}$ ...

Answer 3

Hint: Draw a picture. Use the definition $\sin P=\frac {\text {opposite}}{\text{hypotenuse}}$ You are given all the sides, so . b,c,d are similar. For e, you will have an inverse trig function (there are several choices).

Answer 4

Using definition of trigonometric functions from figure

a) $$\sin \hat{P}=\frac{PQ}{PR}=9/41$$ b) $$\cos \hat{R}=\frac{QR}{PR}=40/41$$ c) $$\csc \hat{P}=\frac{1}{\sin \hat{P}}=41/9$$ d) $$\cot \hat{R}=40/9$$ e) $$\angle \hat{QRP}=\arccos(40/41)$$

Answer 5

You mentioned that $PQR$ is a right triangle. It would be best to draw a picture first.

From here, use the definitions of the trig functions to find parts a through d.

$$a) \sin(P) = \frac{\mbox{opposite side}}{\mbox{hypotenuse}}, \quad b) \cos(R) = \frac{\mbox{adjacent side}}{\mbox{hypotenuse}},$$ $$c) \csc(P) = \frac{\mbox{hypotenuse}}{\mbox{opposite side}}, \quad d) \cot(R) = \frac{\mbox{adjacent side}}{\mbox{opposite side}}.$$

For part e, you can find $\angle R$ using the inverse cosine function, $cos^{-1}$ (also called $\arccos$) with the value you found for $\cos(R)$ in part b.