Transcript Sine Rule File

Trigonometry
Sine Rule
R
B
c
A
q
a
b
C
P
r
Mr Porter
p
Q
Definition: The Sine Rule
In any triangle ABC
a
b
c
=
=
sin A sin B sinC
‘The ratio of each side to the sine of the opposite angle is CONSTANT.
R
B
c
A
q
a
b
C
P
r
p
Q
For triangle ABC
For triangle PQR
a
b
c
=
=
sin A sin B sinC
p
q
r
=
=
sin P sinQ sin R
Example 1: Use the sine rule to find the
value of x correct to 2 decimal places.
C
a
x
b
Example 2: Find the size of α in ∆PQR in
degrees and minutes.
When to apply the sine rule:
When to apply the sine rule:
Is there 2 sides and 2 angles
(or more)?
YES, then use the Sine Rule.
P
118°2
7’
Is there 2 sides and 2 angles
(or more)?
YES, then use the Sine Rule.
q
α
Q
81
A °
c
43
°
B
Write down the sine rule for this triangle
a
b
c
=
=
sin A sin B sinC
Label the triangle.
We do not need the ‘C’ ratio.
Substitute A, a, B and b.
x
8.3
=
sin 81° sin 43
8.3
x=
´ sin 81°
sin 43
Rearrange to make x the subject.
Use calculator.
x = 12.02 (2dec. pl.)
R
p 22.4 cm
Write down the sine rule for this triangle
sin P sinQ sin R
=
=
p
q
r
We do not need the ‘Q’ ratio.
Label the triangle.
Substitute P, p, R and r.
sin a sin118°27'
Rearrange to make sinα the
=
subject.
14
22.4
sin118°27'
sin a =
´14 Evaluate RHS
22.4
sin a = 0.5495 (4dec. pl) To FIND angle, use sin-1 (..)
a = sin -1 ( 0.5495 )
Convert to deg. & min.
a = 33.3327°
a = 33°20'
Ambiguous Case – Angles (Only)
But, could the triangle be drawn a different
way? The answer is YES!
Example 1 : Use the sine rule to find the size
C
of angle θ.
a
When to apply the sine rule:
b
Is there 2 sides and 2 angles
(or more)?
YES, then use the Sine Rule.
θ
θ
41°
c
A
A
B
Write down the sine rule for this triangle to find an angle
sin A sin B sinC
=
=
a
b
c
We do not need the ‘C’ ratio.
Label the triangle.
Substitute A, a, B and b.
sinq sin 41 Rearrange to make sin θ the subject.
=
14.5
9.8
sin 41°
sinq =
´ 14.5 Evaluate RHS
9.8
sinq = 0.9707 (4dec. pl)
To FIND angle, use sin-1 (..)
q = sin -1 ( 0.9707 )
q = 76°6'
C
θ
A
41°
B
By supplementary angles:
q = 180°- 76°6'
q = 103°54 '
Which is correct, test the angle sum to 180°,
to find the third angle, α.
Case 1: a = 180°- ( 76°6'+ 41)
a = 62°54 ' , which is allowed.
Case 2: a = 180°- (103°54 '+ 41)
a = 35°6' , which is allowed.
Hence, both answers are correct!
q = 103°54' or q = 73°6'
Ambiguous Case – Angles (Only)
But, could the triangle be drawn a different
way? The answer is NO!
Example 1 : Use the sine rule to find the size
of angle θ.
R
θ
When to apply the sine rule:
Is there 2 sides and 2 angles
(or more)?
YES, then use the Sine Rule.
p
P
122 Q
8.5 cm °
Write down the sine rule for this triangle
sin P sinQ sin R
=
=
p
q
r
We do not need the ‘P’ ratio.
r
Label the triangle.
Substitute Q, R, q and r.
sinq sin122° Rearrange to make sin θ the subject.
=
8.5
17
sin122°
sinq =
´ 8.5 Evaluate RHS
17
sinq = 0.4240 (4dec. pl)
To FIND angle, use sin-1 (..)
q = sin
-1
( 0.4240 )
q = 25°5'
Lets check the supplementary angle METHOD.
q = 180°- 25°5'
q = 154°55'
Which is correct, test the angle sum to 180°,
to find the third angle, α.
Case 1: a = 180°- ( 25°5'+ 122°)
a = 32°55' , which is allowed.
Case 2:a = 180°- (154°55'+ 122° )
a = -96°55' , which is NOT allowed.
as the triangleangle sum would be
122° + 154°55'+ 96°55' = 373°50'
Hence, the ONLY answer is correct!
q = 32°55'
Example 3: Points L and H are two lighthouses 4 km
apart on a dangerous rocky shore. The shoreline (LH) runs
east–west. From a ship (B) at sea, the bearing of H is
320° and the bearing of L is 030°.
a) Find the distance from the ship (B) to the lighthouse (L),
to the nearest metre.
b) What is the bearing of the ship (B) from the lighthouse
at H?
N
N
0
°
H
0
°
L
4 km
50
°
60
°
70
°
N
40 0 30°
°°
50
°
320°
30
°
4000 m
H
l
Use basic alternate angles in parallel line, bearing and
angle sum of a triangle to find all angles with the ∆BHL.
Re-draw diagram for clarity
60
°
h
d
When to apply the sine rule:
Is there 2 sides and 2 angles
(or more)?
YES, then use the Sine
Rule.
70
°
(a)
B
Write down the (side) sine rule for this triangle
h
b
l
=
=
sin H sin B sin L
We do not need the ‘L’ ratio.
d
4000
=
sin 50° sin 70
d=
B
(ship)
L
b
50
°
Label the triangle.
Substitute B, H, b and d, (h).
Rearrange to make d the subject.
4000
´ sin 50°
sin 70°
Use calculator.
d = 3261m (nearest metre)
(b) From the original diagram :
Bearing of Ship from Lighthouse H:
H = 90°+50°
H = 140°
Example 4: To measure the height of a hill a surveyor
took two angle of elevation measurements from points X
and Y, 200 m apart in a straight line. The angle of elevation
of the top of the hill from X was 5° and from Y was 8°.
What is the height of the hill, correct to the nearest metre?
T
To find ‘h’, we need
either length BY or TY!
3°82°
x
X
172 8°
5
°
200 m ° Y
h
B
When to apply the sine
rule: Is there 2 sides
and 2 angles (or more)?
YES, then use the Sine
Rule to find TY= x.
Use basic angle sum of a triangle, exterior angle of a triangle
and supplementary angles to find all angles with the ∆AYT
And ∆YTB.
Write down the (side) sine rule for this triangle
t
x
y
=
=
sinT sin X sinY
We do not need the ‘Y’ ratio.
x
200
=
sin 5° sin 3°
Substitute X, T, t and x.
Rearrange to make x the subject.
x=
200
´ sin 5°
sin 3°
Use calculator
Do NOT ROUND OFF!
x = 333.0628 m (4dec.pl.)
To find h, use the right angle triangle ratio’s i.e. sin θ.
h
h
= sin 8°
= sin 8°
x
x
h = x sin 8°
h = x sin 8°
200
h = 333.0628 ´ sin 8° h =
´ sin 5° ´ sin 8°
sin 3°
h = 46.3533
h = 46.3533
Hence, the hill is 46 m high (nearest metre).
Example 5:
A wooden stake, S, is 13 m from a point, A,
on a straight fence. SA makes an angle of 20° with the
fence. If a horse is tethered to S by a 10 m rope, where, on
the fence, is the nearest point to A at which it can graze?
sin 20°
Evaluate RHS
´ 13
10
sin B = 0.4446 (4dec. pl)
sin B =
To FIND angle, use sin-1 (..)
Fence
A
B
C
20
°
10 m
10 m
13 m
B = sin -1 ( 0.4446 )
B = 26°24 '
From the diagram, it is obvious that angle B is Obtuse.
•S
Stake
B = 180 – 26° 24’
B = 153° 36’
Then angle ASB = 182 – (153°36’ + 20) = 6° 24’
Now, apply the sine rule to find the length of AB.
Write down the (side) sine rule for this triangle
1) The closest point to A along the fence, is point B.
Hence, we need to find distance AB.
2) Look at ∆ABS, to use the sine rule, need to find angle ABS or
angle ASB.
Write down the (angle) sine rule for this triangle
sin B sin A sin S
=
=
b
a
s
We do not need the ‘S’ ratio.
sin B sin 20
=
13
10
Substitute A, B, a and b.
Rearrange to make sin B the subject.
s
a
b
=
=
sin S sin A sin B
AB
10
=
sin 6°24 sin 20°
10
AB =
´ sin 6°24
sin 20°
AB = 3.259 m
Example 6:
Q, A and B (in that order) are in a straight
line. The bearings of A and B from Q is 020°T. From a
point P, 4 km from Q in a direction NW, the bearing of A
and B are 112°T and 064°T respectively. Calculate the
distance from A to B.
N
0
°
B
Not to scale!
N
0
°112
P
64
°
°
45
°
44
°
x
N
0
°
48
°
23
°
4 km
d
88
°
y
N
0
45°20
° °
92
°
In ∆POA, write down the (side) sine rule for this triangle
q
p
a
=
=
sinQ sin P sin A
We do not need the ‘P’ ratio.
Substitute Q, A, q and y.
y
4
Rearrange to make y the subject.
=
sin 65° sin 92°
4
calculator
y=
´ sin 56° Use
Do NOT ROUND OFF!
sin 92°
y = 3.6274km (4dec.pl.)
In ∆PAB, write down the (side) sine rule. For this triangle.
A
Q
Use basic alternate angles in parallel line, bearing and
angle sum of a triangle to find all angles in the diagram.
To find ‘d’, we need to work backward using the sine rule,
meaning that we must find either x or y first.
[There are several different solution!]
a
p
b
=
=
sin A sin P sin B
We do not need the ‘A’ ratio.
Substitute P, B, y and d.
d
y
Rearrange to make d the subject.
=
sin 48° sin 44°
y
calculator and y = 3.6274
d=
´ sin 48° Use
Do NOT ROUND OFF!
sin 44°
d = 3.88 km (2dec.pl.)