Transcript TOPIC: TRIGOOMETRY Aim: To Understand the 3 Trigonometric

TOPIC: TRIGONOMETRY
Aim: To be able to name the sides of a right
angled triangle (RAT) given and interior angle.
• Identify the names of the sides of these right angledtriangles given angle k b opposite
a
opposite
c
hypotenuse
c
hypotenuse
a
k
adjacent
a
adjacent
k
b adjacent
c
hypotenuse
k
opposite
b
opposite
b
k
c
hypotenuse
a adjacent
• What can you say about the names of the sides?
• They are named according to the angle under
consideration.
TOPIC: TRIGOOMETRY
Aim: To Understand the 3 Trigonometric Ratios
•
The 3 ratios are Sine, Cosine and Tangent
Opposite Side
Sine Ratio 
,
Hypotenuse
Co sin e Ratio 
Adjacent Side
,
Hypotenuse
SOH for short
CAH for short
Opposite Side
Tangent Ratio 
,
AdjacentSide
•
TOA for short
Using pneumonic, the ratios are SOH, CAH,
TOA
TOPIC: TRIGONOMETRY
Aim: To use Trigonometry to find lengths,
given an interior angle and one side of a RAT
•
Finding the length of an unknown side of
a right angled triangle:
a
26o
7cm (adjacent)
•
(opposite)
Calculate the length of y.
y
(opposite)
•
5m (hypotenuse)
34o
NB: The appropriate ratio is sine, SOH
The appropriate ratio to use is Tangent,
i.e. TOA
Tangent Ratio 
•
•
•
•
•
•
Opposite Side
Adjacent Side
Tan 260 = a/7
a/7 = Tan 260
a = 7 x Tan 260
a = 7 x 0.4877
a =3.41cm (to 2 d.p.)
Sine Ratio 
•
•
•
•
•
Opposite Side
Hypotenuse
Sine 340 = y/5
y/5 = Sine 340
y = 5 x Sine 340
y = 5 x 0.559
y =2.80m (to 2 d.p.)
TOPIC: TRIGONOMETRY
Questions for Students to Attempt.
• Calculate the length of side x
• Calculate the length of y
y (Opposite)
Hypotenuse
10.6m
670
x (Adjacent)
• The appropriate ratio to use is
Cosine, i.e. CAH
Co sin e Ratio 
•
•
•
•
Adjacent Side
,
Hypotenuse
Cos 670 = x/10.6
0.39 = x/10.6
x =10.6 X 0.39
x =4.14m (to 2 d.p.)
Hypotenuse
6.2m
420
• NB: The appropriate ratio is
Sine, i.e. SOH
Opposite Side
Hypotenuse
420 = y/6.2
Sine Ratio 
•
•
•
•
•
Sine
y/6.2 = Sine 420
y = 6.2 X Sine 420
y = 6.2 X 0.669
y =4.19m (to 2 d.p.)
TOPIC: TRIGONOMETRY
Starter
•
•
•
•
•
•
Evaluate the following
Tan 50 = 1.19
Sin 50 = 0.76
Cos-1 0.56 = 55.94
= 88.85
Tan-1 40
Sin-10.50 = 30
TOPIC: TRIGONOMETRY
Aim: To Find an interior angle of a RAT
• To find angle a
32cm (Opposite)
25cm
a Adjacent
• The appropriate ratio to
use is Tangent, i.e. TOA
Opposite Side
Tangent Ratio 
Adjacent Side
•
•
•
•
Tan a = 32/25
Tan a = 1.28
a = Tan -1 1.28
a = 52.00
• To find the size of angle y
50cm
(Opposite)
Hypotenuse
30cm
y0
• NB: The appropriate ratio
is
• Sine , i.e. SOH
Sine Ratio 
Opposite Side
Hypotenuse
• Sine y = 30/50
• Sine y = 0.6
• y = Sine-1 0.6
• y = 36.90, approximately 370
TOPIC: TRIGONOMETRY
Questions for students to attempt.
• Find angle b
Opposite
6cm
• Find angle y
12.4m (Adjacent)
y
Hypotenuse
12cm
Hypotenuse
19.7m
b
• The appropriate ratio to
use is Sine, i.e. SOH
Sine Ratio 
•
•
•
•
Opposite Side
Hypotenuse
Sin b = 6/12
Sin b = 0.5
b = Sin-1 0.5
b = 300
• NB: The appropriate ratio
is Cosine, i.e. CAH
Co sin e Ratio 
•
•
•
•
•
Adjacent Side
,
Hypotenuse
Cos y = 12.4/19.7
Cos y = 0.639
y = Cos-1 0.639
y = 50.280
y is approximately 500
TOPIC: TRIGONOMETRY
Aim: To Use Trigonometry to Solve Problems
• Key words
– Angle of Elevation: is an angle between the
horizontal and an object above it
– Angle of Depression /Descent: is an angle
between the horizontal and an object below it
• Tips: Draw and label your diagram
• Write down the formula to be used and calculate
the answer
TOPIC: TRIGONOMETRY
Aim: To Use Trigonometry to Solve Problems
• An aeroplane is 4500m
from touchdown.
• Its angle of descent is 500
to horizontal. How high is
it above the ground.
• The angle of descent is
outside the right angledtriangle. So we need to find
the angle inside it.
• TOA is the ratio to
use
Opposite Side
Adjacent Side
Tan 400
Tangent Ratio 
500
400
a
Runway
4500m
Touchdown
Ground
•
•
•
•
4500/a =
4500/a = 0.8390
a = 4500 / 0.8390
a = 5363.52m
TOPIC: TRIGONOMETRY
Aim: To Use Trigonometry to Solve Problems
• A boat is due South of a Lighthouse.
• It sails on a bearing 0450 until it is
due east of the lighthouse. If the
boat is now 50km away from the
lighthouse, how far has it sailed.
50km (Opposite)
South
0450
a
Hypotenuse
East
• The appropriate
ratio is SOH
Sine Ratio 
•
•
•
•
Opposite Side
Hypotenuse
Sin 450 = 50/a
0.71 = 50/a
a = 50 / 0.71
a = 70.42km