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 :

{ The Earth's sphere and the tangent from the observation point $O$}

Mathematical analysis

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

( R + h ) 2 = D 2 + R 2

Hence

R 2 + 2 R h + h 2 = D 2 + R 2 h ( 2 R + h ) = D 2 D = h ( 2 R + h )

If R = 6.373 × 1 0 6 m, then the variation of D with h is shown in Figure 9.

Figure 9

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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 A B C in Figure 10 which can be regarded as one half of the equilateral triangle A B D , calculate sin, cos, tan for the angles 3 0 and 6 0 .

Figure 10

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By Pythagoras’ theorem: ( B C ) 2 = ( A B ) 2 ( A C ) 2 = x 2 x 2 4 = 3 x 2 4 so B C = x 3 2

Hence sin 6 0 = B C A B = x 3 2 x = 3 2 sin 3 0 = A C A B = x 2 x = 1 2 cos 6 0 = A C A B = 1 2

  cos 3 0 = B C A B = x 3 2 x = 3 2 tan 6 0 = 3 2 1 2 = 3   tan 3 0 = 1 2 3 2 = 1 3

Values of sin θ , cos θ and tan θ can of course be obtained by calculator. When entering the angle in degrees ( e.g. 3 0 ) 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 sin 3 0 are usually simply SIN 30 or, on some calculators, 30 SIN perhaps followed by =.

Task!

(a) Use your calculator to check the values of sin 4 5 , cos 3 0 and tan 6 0 obtained

 in the previous Task.

(b) Also obtain sin 3 . 2 , cos 86 . 8 , tan 2 8 1 5 . (’ denotes a minute = 1 60 )

(a) 0.7071 , 0.8660 , 1.7321   to 4 d.p.

(b) sin 3 . 2 = cos 86 . 8 = 0.0558 to 4 d.p., tan 2 8 1 5 = tan 28.2 5 = 0.5373 to 4 d.p.

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

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Suppose we wish to find the angles at A and B . Clearly sin A = 3 5 , cos A = 4 5 , tan A = 3 4 so we need to solve one of the above three equations to find A .

Using sin A = 3 5 we write A = sin 1 3 5 (read as ‘ A is the inverse sine of 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 3 5 we might use the following keystrokes:

INV SIN 0.6 =

or

÷ 5 INV SIN =

We find sin 1 3 5 = 36.8 7 (to 4 significant figures).

Key Point 5

Inverse Trigonometric Functions

sin θ = x implies θ = sin 1 x cos θ = y implies θ = cos 1 y tan θ = z implies θ = tan 1 z

(The alternative notations arc sin , arc cos , arc tan are sometimes used for these inverse functions.)

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.

A = cos 1 4 5 = 36.8 7 B = cos 1 3 5 = 53.1 3

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.

A = tan 1 3 4 = 36.8 7 B = tan 1 4 3 = 53.1 3

You should note carefully that sin 1 x does not mean 1 sin x .

Indeed the function 1 sin x has a special name – the cosecant of x , written cosec x . So

cosec x 1 sin x (the cosecant function).

Similarly

sec x 1 cos x (the secant function)

cot x 1 tan x (the cotangent function).

Task!

Use your calculator to obtain to 3 d.p. cosec 38 . 5 , sec 22 . 6 , cot 88.3 2 (Use the sin, cos or tan buttons unless your calculator has specific buttons.)

cosec 38 . 5 = 1 sin 38 . 5 = 1.606 sec 22 . 6 = 1 cos 22 . 6 = 1.083

cot 88.3 2 = 1 tan 88.3 2 = 0.029