1.Calculate the angle x in the triangle below:

Using the Sine Rule:
    a                b      
 Sin A   ⁼     Sin B

    9                6      
 Sin x   ⁼     Sin 34^o

If we "cross-multiply" we get:

6Sin x = 9Sin 34^o

x = sin^-1(9Sin 34^o / 6)

x = 57.01^o

Using the Cosine Rule:

c² = a² + b² - 2abCosC

y² = 5² + 9² - 2(5 x 9)Cos 42^o

y² = 106 - 90Cos42^o

y = Ö (106 - 90Cos42^o)

y = 6.25

3.  In the quadrilateral below, calculate the size of the angle ABC:

Calculate the length of AC first by using the Cosine Rule:

AC² = 5² + 4² - 2(5 x 4)Cos 100^o

AC = Ö (41 - 40Cos100^o)

AC = 6.92

Now use the sine rule to calculate the angle ABC, which we will call x.

 6.92              10      
 Sin x   ⁼     Sin 60^o

Sin x = 6.92Sin 60^o / 10

x = sin^-1(6.92Sin 60^o / 10)

x = 36.85^o

4. The picture below shows a farmer's field after it has been mapped out.
i) Find the length AC
ii) Find the angle ADC
iii) Find the area of the field

i) Use the Cosine Rule:

AC² = 30² + 85² - 2(30 x 85) Cos 104^o

AC = Ö (8125 - 5100Cos104^o)

AC = 96.74m

ii) Let the angle ADC = x and use the Cosine Rule again:

96.74² = 45² + 100² - 2(45 x 100) Cos x

9,358.63 = 12,025 - 9,000 Cos x

-2,666.37 = 9,000 Cos x

Cos x = -2,666.37 / 9,000

x = Cos-1(-0.2962636)

x = 107.23o

iii) The area of a non-right angled triangle is given by the formula (this is worth remembering):
Area = ½ ab Sin C

If you look at the field as being two triangles stuck together, then you can calculate the area of each triangle and then add them together to give the area of the field:

Area of field =
½ (30 x 85) Sin 104^o + ½ (45 x 100) Sin 107.23^o

= 1,275 Sin 104^o + 2,250 Sin 107.23^o

= 1,237.13 + 2,149.03

= 3,386.16m²

5.

Using the details given in the triangle above:

a) Show that 3x² + 6x - 36 = 0
b) Solve the equation 3x² + 6x - 36 = 0 and give your answer to 3 significant figures.
c) D is a point on the line AC such that the angle ADB = 100^o. Calculate the length of BD.

a) At first glance there seems to be no connection between the triangle and the quadratic equation, however, the quadratic has a squared term in and so does the Cosine Rule. As only one angle is given we must be expected to apply the Cosine Rule as follows:

(2x)² = x² + 6² - 2(x x 6) Cos 60^o

If you work out Cos 60^o on your calculator you'll realise that it equals 0.5, this gives us:

4x² = x² + 36 - 6x
3x² = 36 - 6x
3x² + 6x - 36 = 0

b) Use the quadratic formula to solve the equation for x

x = 6 ± Ö (36 - (4 x 3 x -36) / (2 x 3)

x = 22.6 (to 3 s.f.)

c) Using the Sine Rule:

     6                  BD      
Sin 100^o   =    Sin 60^o

BD = 6 Sin 60^o / Sin 100^o

BD = 5.28

6. A hiker starts her journey at point A. She notices a farm house at point C and works out its bearing is at 138^o. She then walks for 5 kilometres and stops at point B. At point B the hiker looks again at the farm house and calculates its bearing now to be 200^o. Calculate the distances AC and BC.

Here is a diagram of the hiker's path and bearing measurements:

First you should calculate the angles in the triangle:

CAB = 138^o - 50^o = 88^o

Calculating ABC is slightly trickier and requires some knowledge of angle properties:

From the properties of angles in parallel lines it can be seen that x = 40^o also because the blue line makes an angle of 90^o with the North line going anticlockwise, the angle from the North to the blue line clockwise = 360^o - 90^o = 270^o. Hence:

ABC = 270^o - 40^o - 200^o = 30^o

All angles in a triangle add up to 180^o, hence ACB = 180^o - 88^o - 30^o = 62^o

Now we know all the angles, we can use the Sine Rule to find distances AC and BC.

     5                   AC   
 Sin 62^o   =     Sin 30^o

AC = 5Sin 30^o / Sin 62^o

AC = 2.83

     5                   BC   
 Sin 62^o   =     Sin 88^o

BC = 5 Sin 88^o / Sin 62^o

BC = 5.66

7.

The diagram above represents air traffic control at point C. An aeroplane follows a direct path from A to B.

i) Calculate the distance the aeroplane flies between the points A and B

ii) At its shortest distance from air traffic control, the aeroplane is at the point X on the line AB. Calculate the distance CX.

i) Use the Cosine Rule to calculate AB:

AB² = 10² + 12² - 2(10 x 12)Cos 140^o

AB = Ö (244 - 240Cos 140^o)

AB = 20.68km

ii) At its shortest distance from C, the point X will be directly under C (i.e. the angle AXC will be 90^o).

So, we have a right-angled triangle and know the length of the hypotenuse, but we don't know an angle.

If we use the Sine rule we can calculate angle CAB (which is the same as CAX).

  20.68                 12   
 Sin 140^o   =     Sin CAB^o

Sin CAB = 12 Sin 140^o  / 20.68

CAB = 21.9^o

Now all we need to do is some simple trigonometry to work out the length of CX.

Sin 21.9^o = CX / 10

CX = 3.73km

8. A ship sails from harbour and travels 25km on a bearing of 30^o before reaching a marker bouy. At this point the ship turns and follows a course on a bearing of 90^o and travels for 32km until it reaches an island. On the return journey, the ship is able to take the most direct route back to the harbour.

i) Draw a diagram to represent both the outbound and return journeys.

ii) What is the total distance travelled by the ship?

i)

Note that the properties of angles and parallel lines had to be used again to determine the angle of 120^o.

ii) We can now use the Cosine Rule to determine the length IH:

IH² = 25² + 32² - 2(25 x 35) Cos 120^o

IH = Ö (574 - 1750 Cos 120^o)

IH = 38km

Hence the total distance travelled by the ship is

25 + 32 + 38 = 95km