Question

The picture shows a triangular island:

Answer 1

Answer:

1. $q=\sqrt{r^{2}+s^{2}}$

2. $q=\frac{s}{\cos 55}$ and $q=\frac{r}{\cos 55}$

Step-by-step explanation:

We are given that,

A right triangle with acute angle 55° having the sides lengths as shown in the figure below.

1) Using the Pythagoras Theorem, which gives the result,

$Hypotenuse^{2}=Perpendicular^{2}+Base^{2}$

So, we have the equation from the figure,

$q^{2}=r^{2}+s^{2}$

i.e. $q=\sqrt{r^{2}+s^{2}}$

Thus, the expression showing the value of q is $q=\sqrt{r^{2}+s^{2}}$.

2) Using the trigonometric forms in a right triangle, we have,

$\cos x=\frac{Adjacent}{Hypotenuse}$

i.e. $\cos 55=\frac{s}{q}$

i.e. $q=\frac{s}{\cos 55}$

Or $\sin x=\frac{Opposite}{Hypotenuse}$

i.e. $\sin 55=\frac{r}{q}$

i.e. $q=\frac{r}{\cos 55}$

The expressions for the value of q are $q=\frac{s}{\cos 55}$ and $q=\frac{r}{\cos 55}$

Answer 2

Answer:

The expressions that show the value of q are

1) $q=\sqrt{r^{2}+s^{2}}$

2) $q=\frac{s}{cos(55\°)}$

3) $q=\frac{r}{sin(55\°)}$

4) $q=\frac{s}{sin(35\°)}$

5) $q=\frac{r}{cos(35\°)}$

Step-by-step explanation:

see the attached figure to better understand the problem

we know that

case A)

In the right triangle of the figure

Applying the Pythagoras Theorem

$q^{2}=r^{2}+s^{2}$

$q=\sqrt{r^{2}+s^{2}}$

case B)

In the right triangle of the figure

$cos(55\°)=\frac{s}{q}$

solve for q

$q=\frac{s}{cos(55\°)}$

case C)

In the right triangle of the figure

$sin(55\°)=\frac{r}{q}$

solve for q

$q=\frac{r}{sin(55\°)}$

case D)

In a right triangle

if $A+B=90\°$

then

$cos(A)=sin(B)$

therefore

$q=\frac{s}{cos(55\°)}$------> $q=\frac{s}{sin(35\°)}$

$q=\frac{r}{sin(55\°)}$ ------>  $q=\frac{r}{cos(35\°)}$