1.

The diagram shows the cross-section of a house. The tip of the house roof is directly central.

i) Calculate the length of the slope of the roof BC.

ii) Explain two methods which can be used to determine the angle of slope x^o.

iii) Use one of your methods to find the angle x^o.

Give all answers to 3 significant figures.

i) As the house roof is directly central you can use Pythagoras' Theorem to work out BC where one side = 7m and the other = 12.5m (i.e. a half of 25m).

so BC² = 7² + (12.5)²

BC = Ö(7² + (12.5)²

BC = Ö(205.25)

BC = 14.3m (to 3 s.f.)

ii) sin x = 7 / BC or tan x = 7 / 12.5 (you can also use cos x = 12.5 / BC)

iii) In these sort of "follow-on" questions it's a good idea to avoid using the value you've just calculated if at all possible. If the value you've just calculated for BC is wrong and you use BC in the calculation here then you stand to lose a mark for getting the wrong answer for this question (you'll still get the method marks though).

So, I'd choose tan x = 7 / 12.5

x = tan^-1(7 / 12.5) = 29.2^o (to 3 s.f.)

sin 45^o = x / 2.25

Hence (with a smattering of algebra):

x = 2.25 sin 45^o

x = 1.59m

3.  

In a computer game, when the mouse arrow is clicked, the puppy will run towards it. The movement of the puppy is described in the computer program as being a horizontal movement followed by a vertical movement. The positions of the puppy and mouse are shown in the picture as (horizontal movement, vertical movement).

To reach the arrow, at what angle from the horizontal must the puppy travel? (Give your answer to 2 d.p.)

The horizontal distance travelled will be 65 - 20 = 45

The vertical distance travelled will be 108 - 30 = 78

Surprise, surpise, this makes up most of a right-angled triangle, so:

tan x = 78 / 45

x = tan^-1 (78 / 45)

x = 60.02^o.

4. 

The diagram shows a bridge (not drawn to scale) with the following measurements:

AC = 60m
BC = 80m
AF = 300m
BD = 150m
DE = 40m

Calculate the distance AB and hence calculate the angle X.

As you are told that AC = 60m and BC = 80m, you can calculate AB using Pythagoras' Theroem:

Note that BC is the hypotenuse, so:

80² = AB² + 60²

AB² = 80² - 60²

AB = Ö(80² - 60²)

AB = 52.92m (to 2 d.p)

You know that AF = 300m and now that AB =100m, you are also told that BD = 150m.

So, DF = 300 - 100 - 52.92 = 147.08m

Now you know DF and are told that DE is 40m you can calculate x:

sin x = 40/147.08

x = sin^-1(40/147.08)

x = 15.78^o (to 2 d.p.)

5.

Barnacle Bob looks over the cliff edge at a boat on the sea. He is able to measure the angle of despression as 50^o. What is the distance of the boat from the cliff?

To find the angle in the triangle, you need to subtract 50^o from 90^o (the angle the vertical line of the cliff makes with the horizontal line from which the 50^o is measured).

90^o - 50^o = 40^o.

Now you have a simple right-angled triangle problem to solve. As you are not given the length of the hypotenuse, you have to use tan. Let the length from the boat to the cliff = x.

tan 40^o = x / 60

x = 60 tan 40^o

x = 50.35m (to 2 d.p.)

6. Mountain rescue set off from base and walk 8km then 6km west before finding the injured climber. They now need to find the quickest way back to base. What bearing should they take (to the nearest degree) ?

Here is the picture you should draw to understand what is happening:

The quickest path back to base happens conveniently to be along the hypotenuse of the right-angled triangle (shown by the dotted blue line).

Bearings are always measured from the North (indicated by the red arrow. Hence the bearing will be 90^o + the angle in the triangle (shown by the green arrow). Now all we need to do is find the angle in the triangle, we'll calll this x:

tan x = 8 / 6

x = tan^-1 (8 / 6)

x = 53.13^o (to 2.d.p)

Now, remember to add on the 90^o, to give the bearing from the vertical North line:

Bearing = 90^o + 53.13^o = 143.13^o.

7.

The diagram shows a regular prism with a right angled triangle as its cross-section. The midpoint of line CD is indicated by M. Find the following lengths and angles:

i) AC
ii) EC
iii) ECF
iv) FC
v) ACB
vi) EMF

i) Use Pythagoras:
AC² = 4² + 10²
AC = Ö(16 + 100)
AC = Ö116
AC = 10.77

ii) We can use Pythagoras again, note that AF = BE as this is a regular prism.

EC² = 15² + 10²
EC = Ö(225 + 100)
EC = Ö325
EC = 18.03

iii) Angle ECF is the angle between EC and CF. We have just calculated EC as 18.03 and we know the length of EF is the same as AB i.e. 4 (regular prism). So we have a right angled triangle again:

tan ECF = EF / EC
tan ECF = 4 / 18.03
ECF = tan^-1 (4 / 18.03)
ECF = 12.51^o.

iv) sin 12.51 = EF / FC

So, sin 12.51 = 4 / FC

Using a bit of algebra:
FC sin 12.51 = 4
FC = 4 / sin 12.51
FC = 18.47

v) Angle ACB can be calculated easily:

tan ACB = 4 / 10
ACB = tan^-1 (4 / 10)
ACB = 21.80^o.

vi) Angle EMF is a bit trickier. M is the midpoint of CD, hence DM = half of CD = 7.5. Notice that we can calculate EM using Pythagoras:

EM² = DE² + DM²
EM² = 10² + (7.5)²
EM = Ö(100 + 56.25)
EM = 12.50

Now we know EM = 12.5 and we know EF = 4, we can calculate angle EMF using tan.

tan EMF = 4 / 12.5

EMF = tan^-1( 4 / 12.5)

EMF = 17.74^o.

8.

The diagram shows a cuboid. Find the length of the diagonal BH, when BC = 1cm, CD = 2cm and DH = 3cm.

You can use Pythagoras' Theorem in 3d to solve this!

BH² = BC² + CG² + DH²

BH = Ö(BC² + CG² + DH²)

BH = Ö(1² + 2² + 3²)

BH = Ö14

BH = 3.74cm