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Mathematics and statistics, Questions and answers
Typology: Exercises
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Thomas Whitham Sixth Form S J Cooper
Using Pythagoras’ Theorem
c cm
c
Using Pythagoras’ Theorem
y cm
y
96 9. 80
If it is a calculator paper there is no need for this line as the above can be typed into the calculator.
Here we have a difference of the squares as we are finding one of the smaller sides.
Example
Find the exact length of AB X distance from A to B = 6 Y distance from A to B = 2
First The hypotenuse (faces the right angle always). Next The side Opposite to your angle Then The Adjacent side (the last side)
Opposite
Adjacent
Hypotenuse
A
Which enables you to write down TOA / SOH / CAH Leading to the three trig ratios:
Adjacent
tan A Opposite Hypotenuse
sin A Opposite Hypotenuse
cos A Adjacent
Solution Firstly identify the sides AC is the hypotenuse BC Opposite angle A AB is Adjacent to A. Hence we require the Tan ratio i.e. tan A OppositeAdjacent
tan^5
Example Calculate the size of angle A in the triangle ABC below
Solution Labelling the sides we are given the Hypotenuse and the Opposite. So we use the Sine ratio
A
C
8 cm^ B
5 cm
On our calculators we work out 5 8. Then press “inverse tan” i.e. tan^1. Which is usually shift and tan buttons.
A
C
6 cm B
4 cm
XY cm
cos 41
cos 41 4. 6
cos 41 4.^6
Example
Example
Here use our algebra skills and rearrange the equation above by multiplying both sides by XY. And then divide both sides by cos 41.
8cm 8cm 8cm 8cm
6 cm 3 cm 3 cm
70°
70° (^) 8 cm
70°
70° 4 cm 4 cm
48°
24° 24°
P
Q R
P
Q R
Example To calculate the size of the angle PQR in the a foregoing triangle PQR
Solution Concentrate on one right angled triangle Here we know the hypotenuse and adjacent
cos^3
Example Find the length of AB in the triangle below
Solution Start by finding the length x in the Right angled triangle formed Here we are given the hypotenuse and want to find the opposite, therefore
5 sin 24
sin (^245)
x
x
x
8cm 8cm
3 cm 3 cm
P
Q R
48°
A
C
B
24°
A
C
B
x
5 cm
5 cm
5 cm
5 cm
Example
Find c in the triangle below
𝑐 sin 50 =^
7 sin 60
𝑐 = (^) sin 60^7 × sin 50 𝑐 = 6.19cm
Example Find a in the triangle below
𝑎 sin 27 =^
8 sin 110
𝑎 = (^) sin 110^8 × sin 27 𝑎 = 3.87cm
Example Find x in the triangle below
sin 40 =^
sin 86
𝑥 = (^) sin 86 10.6 × sin 40 𝑥 = 6.83cm
Example
Find the size of angle B in the triangle below
Here we use the sine rule writing it upside down! An additional twist – we must first find angle C, as we don’t know the length opposite angle B. sin 𝐶 6 =
sin 120 15 sin 𝐶 = sin 120 15 × 6 sin 𝐶 = 0.346.. 𝐶 = 20.3°
First we need to work out the angle opposite the 10.6cm. Using angles in a triangle. Angle = 86
Using the cosine rule:
𝑏^2 = 5^2 + 8^2 − 2 × 5 × 8 × cos 112
𝑏^2 = 118.
𝑏 = 10.91cm
Example
Find the length x in the triangle below
𝑥^2 = 10 + 9^2 − 2 × 10 × 9 × cos 64 𝑥^2 = 102. 𝑥 = 10.1cm
Example
Find the length y in the triangle below
𝑦^2 = 12.3^2 + 7.5^2 − 2 × 12.3 × 7.5 × cos 48 𝑦^2 = 84. 𝑦 = 9.17cm
which can be typed into the calculator in one go!
Example Find angle C
Here we could use the cosine rule and rearrange to find Angle C. Or use the rearranged cosine rule which is cos 𝐴 = 𝑏 (^2) +𝑐 2 −𝑎 2 2 𝑏𝑐 62 = 7^2 + 5^2 − 2 × 7 × 5 × cos 𝐴 36 = 74 − 70 cos 𝐴 70 cos 𝐴 = 38 cos 𝐴 =^3870 𝐴 = 57.1°
Example Find angle P
Using the rearranged formula
cos 𝑃 =^11
cos 𝑃 =^149176 𝑃 = 32.2°
(i) Using the Sine rule we can find AB 𝐴𝐵 sin 72 =^
sin 63 𝐴𝐵 = (^) sin 63^5 × sin 72 𝐴𝐵 = 5.34cm (ii) Area = 12 × 5 × 5.34 × sin 45 = 9.43cm^2
Example
Find the area of triangle PQR
Using the cosine rule – find an angle inside the triangle
92 = 6^2 + 4^2 − 2 × 6 × 4 × cos 𝑃 81 = 52 − 48 cos 𝑃 48 cos 𝑃 = − 29 cos 𝑃 = − (^2948) 𝑃 = 127.2°
Area = 12 × 6 × 4 × sin 127. = 9.56cm^2
75 tan 43
tan (^4375)
x
x
x
height = 69.94 + 50 = 119.94 m
Using the right angled triangle given
A
B
43°
50m
75m
43° 75m
x
(ii) Using the triangles on the previous page we can find the angle of inclination using right angled trigonometry.
tan^3
tan^5
Hence the angle of inclination for path AD = 36.87°, and for path DB = 26.57°
Example On a radar two planes A and B can be seen on bearings 075° and 154°, respectively and at distances 15km and 7km from the control tower.
Find the distance between the two planes.
75 79
15km
7km
B
A
N
Using the cosine rule
AB km
(^215 2722157) cos 79
C D or C 2 r Arc length 360
l D
Perimeter of a circle is called the circumference. Example Work out the circumference of the circle below
cm
Example What is the length of fencing required to enclose a semi-circular field with diameter 80m. Give your answer correct to 1 decimal place.
