Trigonometry.

GCSE MATHEMATICS Intermediate Tier, topic sheet.
S O HC A HT O A
1. Calculate the size of angle ABC in the following diagram.
C
8 cm
10 cm
A
B
2. This diagram shows the design for a sail. Angles ABC and ADB are right angles. BC is 5
metres long and BD is 2 metres long.
A
a) Calculate the length of DC.
b) Calculate the size of angle CBD.
c) Calculate the length of AD.
D
2 m
B
d) What is the length of AC.
5 m
C
3.
C
8 cm
D
4 cm
B
10 cm
A
The diagram shows a sail ABC with a support rod AD. The length of AB is 10 cm and the
distances BD and CD are 4 cm and 8 cm respectively. Angle ABD is 90 o .
Calculate the size of angle CAD.
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4. a) The angle of elevation of the top of a building from a point 75 m horizontally from
the foot of the building is 48 o . Calculate the height of the building.
b) Triangle PQR is right-angled at R, the length of PQ is 35 m and the length of QR is
12 m. Calculate the size of QPR ˆ .
Q
35 m
12 m
P
R
5. Calculate the size of angle CAD in the diagram below.
C
D
16 cm
12 cm
B
8 cm
A
6. In the diagram below, angle BAC = 60 o , CD = 12 cm and AC = 10 cm.
Calculate the length of BD.
B
D
12 cm
C
10 cm
60 o
A
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SOLUTIONS / ANSWERS.
REMEMBER THAT S O HC A HT O A ONLY
APPLIES TO RIGHT-ANGLED
TRIANGLES.
1. First label the sides with Opp (for Opposite), Hyp (for Hypotenuse) and Adj (for Adjacent)
etc.
C
Opp = 8 cm
Hyp = 10 cm
Angle ABC
A
B
Since we are dealing with O and H, we use the Sin button on the calculator.
Sin ABC = H
O = 10
8 .
{always follow the Sin, Cos, Tan with the angle!}
So, Sin ABC = 10
8 .
There is no point trying to press the Sin button since we do not know the angle ABC.
Instead we use Inv Sin
So, push Inv Sin 10
8 to get the answer ABC = 53.13 o .
{use the fraction button a
b c }
2. a) Use the Pythagoras theorem in the right-angled triangle BDC.
5 2 = 2 2 + DC 2
25 = 4 + DC 2
21 = DC 2
DC = 4.582575695 m {don’t round up because you’ll probably need this answer
again.}
b) Again, use the right-angled triangle BDC.
A 2
Cos CBD =
H = 5
CBD = 66.42182152 o .
D
Adj = 2 m
B
Hyp = 5 m
angle CBD
{after pressing Inv Cos etc.}
C
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c) Now use the right-angled triangle ABD.
A
Opp
90 66.42182152 o = 23.57817848 o
D
Adj = 2 m
B
Tan 23.57817848 o O
=
A =
which means 0.43643578 =
AD
2
AD .
2
Thus, after multiplying by 2 we get AD = 0.872871561 m.
d) AC = AD + DC = 0.872871561 + 4.582575695 = 5.455447256 m.
3.
C
angle CAD
8 cm
D
4 cm
B
10 cm
A
Since the required angle is NOT part of a right-angled triangle we must work around the
problem.
First consider triangle ABC.
C
Opp = 12 cm
Tan x =
O
A =
12
10
{Inv Tan}
x = 50.19442891 o .
B
Adj = 10 cm
x
A
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Now consider triangle ABD.
D
Opp = 4 cm
B
Adj = 10 cm
y
A
Tan y =
O
A =
4
10
{Inv Tan}
y = 21.80140949 o .
To finish off, simply note that angle CAD = x y = 50.19442891 21.80140949
= 28.39301942 o .
4. a) You must first draw a diagram for this.
Building
48 o
75 m
Using Tan: tan 48 o O
=
A =
which means 1.110612515 =
height
75
height .
75
Multiplying by 75 gives
height = 83.30 m {2 decimal places}.
O 12
b) Using Sin: sin QPR =
H = 35
{Inv Sin}
QPR = 20.05 o {2 decimal places}.
Q
Hyp = 35 m
Opp = 12 m
P
R
5. This is very similar to question 3) but uses Inv Cos instead of Inv Tan.
Answer = 11.8 o {1 decimal place}.
6. Use S O HC A HT O A in the right-angled triangle ABC to get BC = 17.32050808 cm.
This means that BD = 17.32050808 12 = 5.3 cm {1 decimal place}.
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