3 Engineering Example 2

3.1 Horizon distance

Problem in words

Looking from a height of 2 m above sea level, how far away is the horizon? State any assumptions made.

Mathematical statement of the problem

Assume that the Earth is a sphere. Find the length $D$ of the tangent to the Earth’s sphere from the observation point $O$ .

Figure 8 :

Mathematical analysis

Using Pythagoras’ theorem in the triangle shown in Figure 8,

${\left(R+h\right)}^2=D^2+R^2$

Hence

$R^2+2Rh+h^2=D^2+R^2\to h\left(2R+h\right)=D^2\to D=\sqrt{h\left(2R+h\right)}$

If $R=6.373\times 10^6$ m, then the variation of $D$ with $h$ is shown in Figure 9.

Figure 9

At an observation height of 2 m, the formula predicts that the horizon is just over 5 km away. In fact the variation of optical refractive index with height in the atmosphere means that the horizon is approximately 9% greater than this.

Task!

Using the triangle $ABC$ in Figure 10 which can be regarded as one half of the equilateral triangle $ABD$ , calculate sin, cos, tan for the angles $30^{\circ }$ and $60^{\circ }$ .

Figure 10

Answer

Values of $sin\theta ,cos\theta$ and $tan\theta$ can of course be obtained by calculator. When entering the angle in degrees ( e.g. $30^{\circ }$ ) the calculator must be in degree mode. (Typically this is ensured by pressing the DRG button until ‘DEG’ is shown on the display). The keystrokes for $sin30^{\circ }$ are usually simply SIN 30 or, on some calculators, 30 SIN perhaps followed by =.

Task!

(a) Use your calculator to check the values of $sin45^{\circ },cos30^{\circ }$ and $tan60^{\circ }$ obtained

 in the previous Task.

(b) Also obtain $sin3.2^{\circ },cos86.8^{\circ },tan28^{\circ }15^{\prime }$ . (’ denotes a minute $={\frac{1}{60}}^{\circ }$ )

Answer

3.2 Inverse trigonometric functions (a first look)

Consider, by way of example, a right-angled triangle with sides 3, 4 and 5, see Figure 11.

Figure 11

Suppose we wish to find the angles at $A$ and $B$ . Clearly $sinA=\frac{3}{5},cosA=\frac{4}{5},tanA=\frac{3}{4}$ so we need to solve one of the above three equations to find $A$ .

Using $sinA=\frac{3}{5}$ we write $A={sin}^{-1}\left(\frac{3}{5}\right)$ (read as ‘ $A$ is the inverse sine of $\frac{3}{5}$ ’)

The value of $A$ can be obtained by calculator using the ‘ ${sin}^{-1}$ ’ button (often a second function to the $sin$ function and accessed using a SHIFT or INV or SECOND FUNCTION key).

Thus to obtain ${sin}^{-1}\left(\frac{3}{5}\right)$ we might use the following keystrokes:

INV SIN 0.6 =

or

3  $÷$ 5 INV SIN =

We find ${sin}^{-1}\frac{3}{5}=36.87^{\circ }$ (to 4 significant figures).

Task!

Check the values of the angles at $A$ and $B$ in Figure 11 above using the ${cos}^{-1}$ functions on your calculator. Give your answers in degrees to 2 d.p.

Answer

Task!

Check the values of the angles at $A$ and $B$ in Figure 11 above using the ${tan}^{-1}$ functions on your calculator. Give your answers in degrees to 2 d.p.

Answer

You should note carefully that ${sin}^{-1}x$ does not mean $\frac{1}{sinx}$ .

Indeed the function $\frac{1}{sinx}$ has a special name – the cosecant of $x$ , written $\mathrm{\text{cosec}}x$ . So

$\mathrm{\text{cosec}}x\equiv \frac{1}{sinx}$ (the cosecant function).

Similarly

$secx\equiv \frac{1}{cosx}$ (the secant function)

$cotx\equiv \frac{1}{tanx}$ (the cotangent function).

Task!

Use your calculator to obtain to 3 d.p. $\mathrm{\text{cosec}}38.5^{\circ },sec22.6^{\circ },cot88.32^{\circ }$ (Use the sin, cos or tan buttons unless your calculator has specific buttons.)

Answer