Transcript Unit 3x

CHAPTER 3
Acute Triangle Trigonometry
3.1 – Exploring Side-Angle Relationships in
Acute Triangles
Chapter 3: Acute Triangle Trigonometry
Trigonometry
What are the three trigonometric functions?
Sine (sin), Cosine (cos), and Tangent (tan)
Remember:
SOH
CAH
TOA
Hypotenuse
Opposite
sin
Hypotenuse
Adjacent
cos
Adjacent
Opposite
tan
Example
Solve for height, h:
sin(46.5) = h/51.2cm (sin=opp/hyp)
 h = 51.2 × sin(46.5)
 h = 37.1 cm
Or:
sin(83.3) = h/37.3cm
 h = 37.3 × sin(83.3)
 h = 37.1 cm
So:
37.3 × sin(83.3) = 51.2 × sin(46.5)
Or:
37.3
51.2
=
sin(46.5) sin(83.3)
• What’s the relationship
between 37.3 cm and 46.5°?
• How about 51.2 cm and 83.3°?
• What would have happened if
we had put the h going from
vertex B down to its opposite
side? What could we say
about 39.7 cm and 50.4°?
The Sine Law
Law of
Sines:
OR
a
b
c
=
=
sin A sin B sinC
PG. 131, # 1-5
Independent Practice
3.2 – Proving and Applying the Sine Law
Chapter 3: Acute Triangle
Last time we learned the Sine Law:
Proof:
Draw a triangle like the last, with another perpendicular
line from b to vertex B, called h2.
h2
Remember!
sin = opp/hyp
sin(C) = h/b
sin(B) = h/c
sin(A) = h2/c
sin(C) = h2/a
h = b x sin(C) = c x sin(B)
 b = c .
sin(B) sin(C)
h2 = c x sin(A) = a x sin(C)
 c = a .
sin(C) sin(A)
a
b
c
=
=
sin A sin B sinC
Example
A triangle has angles measuring 80° and 55°. The side opposite the 80° angle is 12.0 m
in length. Determine the length of the side opposite the 55°angle to the nearest tenth of
a metre.
Draw a diagram:
What will the third
angle be? What’s the
total sum of angles in
a triangle?
Which side are
we looking for?
Sine Law:
a
b
c
=
=
sin A sin B sinC
12.0
b
=
sin 80 sin 55
sin 55 (
To the nearest tenth of a metre, the side
opposite the 55° angle is 10.0 m long.
12.0
)=b
sin 80
b = 9.981...
Example
Toby uses chains attached to hooks on the ceiling and a winch to lift engines at his father’s
garage. The chains, the winch, and the ceiling are arranged as shown. Toby solved the
triangle using the sine law to determine the angle that each chain makes with the ceiling to
the nearest degree. He claims that θ=40° and α=54°. Is he correct? Explain, and make
any necessary corrections.
2.2
2.8
=
sinq sin 86
Þ 2.2 ´ sin86 = 2.8 ´ sinq
Þ
So, if θ=51°, how can we tell
what α is? Remember, the sum
of the angles must be 180°.
2.2 ´ sin 86
= sin q
2.8
Þ sinq = 0.7838...
Þ q = sin-1 (0.7838...) = 51.396
Þq
51
θ+α+86° = 180°
51°+α+86° = 180°
α=180°-51°-86°=43°
So, Toby was incorrect, in actuality θ=51° and α=43°.
Example
The captain of a small boat is delivering supplies to two lighthouses, as shown. His compass
indicates that the lighthouse to his left is located at N30°W and the lighthouse to his right is
located at N50°E. Determine the compass direction he must follow when he leaves lighthouse
B lighthouse A.
sin B sinC
=
b
c
sin B sin 80
=
9
12
sin B = 9(
sin 80
)
12
sin B = 0.7386...
-1
ÐB = sin (0.7386...)
ÐB = 47.612
The angle B is 47.612°,
but we want the direction
he’d go from lighthouse B.
We need to draw a line
pointing East from vertex B.
O
Since the new line goes West to
East, it must be perpendicular to
the line going South to North…
so the angle is 90°.
If we know that one angle
in ΔBCO is 90° and another
is 50°, how can we figure
out the third angle?
∠CBO = 180° - 50° - 90°
= 40°
∠NBO = 47.6° - 40° = 7.6°
To find the direction on the
compass, we must subtract our
answer from 90°. So, he will
head N82°W from lighthouse B.
PG. 138-141, # 2-15, 18
Independent Practice
3.3 – Proving and Applying the Cosine Law
Chapter 3: Acute Triangle Trigonometry
Cosine Law
In any acute triangle,
a2 = b2 + c2 – 2bc cosA
b2 = a2 + c2 – 2ac cosB
c2 = a2 + b2 – 2ab cosC
Proof of the Cosine Law
In any acute triangle,
a2 = b2 + c2 – 2bc cosA
b2 = a2 + c2 – 2ac cosB
c2 = a2 + b2 – 2ab cosC
Proof:
h2
c2
x2
= –
h2 = b2 – y2
 c2 – x2 = b2 – y2
 c2 = x2 + b2 – y2
 c2 = (a – y)2 + b2 – y2
(since x = a – y)
 c2 = a2 – 2ay + y2 + b2 – y2
 c2 = a2 + b2 – 2ay
cosC = y/b
b cosC = y
(cos = adj/hyp)
c2 = a2 + b2 – 2ay
c2 = a2 + b2 – 2ab cosC
The other parts can be
proven similarly.
Example
Determine the length of CB to the nearest metre.
a2 = b2 + c2 – 2bc cosA
a2 = 322 + 402 – 2(32)(40)cos(58°)
a2 = 1024 + 1600 – 2560cos(58°)
a2 = 2624 – 2560cos(58°)
a2 = 1267.406
a = 35.600…
Why can’t you use the
Sine Law?
CB is 36 m.
Example
A three-pointed star is made up of an equilateral triangle and three congruent isosceles
triangles. Determine the length of each side of the equilateral triangle in this three-pointed
star. Round the length to the nearest centimetre.
First, label the diagram!
x2 = z2 + y2 – 2zy cosX
x2 = 602 + 602 – 2(60)(60)cos(20°)
x2 = 3600 + 3600 – 6765.786…
x2 = 434.213
x = 20.837…
Each side of the
equilateral triangle
has a length of 21 cm.
PG. 150 – 153, #1, 4, 5, 7,
8, 10, 11, 13, 15.
Independent Practice
3.4 – Solving Problems Using Acute Triangles
Chapter 3: Acute Triangle Trigonometry
Example
Two security cameras in a museum must be adjusted to monitor a new display of fossils. The
cameras are mounted 6 m above the floor, directly across from each other on opposite
walls. The walls are 12 m apart. The fossils are displayed in cases made of wood and
glass. The top of the display is 1.5 m above the floor. The distance from the camera on the
left to the centre of the top of the display is 4.8 m. Both cameras must aim at the centre of
the top of the display. Determine the angles of depression, to the nearest degree, for each
camera.
sinθ=4.5/4.8
θ=sin-1(4.5/4.8)
Draw a diagram:
θ=69.635…°
Example continued…
z2 = d2 + b2 – 2db cosA
z2 = 122 + 4.82 – 2(12)(4.8) cos(69.635)
z2 = 126.952…
z = 11.267…
sinα= 4.5/11.267
α= sin-1(4.5/11.267)
α= 23.539°
Camera A must be adjusted to an angle of 70° and
camera B must be adjusted to an angle of 24°.
Example
Brendan and Diana plan to climb the cliff at Dry Island Buffalo Jump, Alberta. They need to
know the height of the climb before they start. Brendan stands at point B, as shown in the
diagram. He uses a clinometer to determine ∠ABC, the angle of elevation to the top of the
cliff. Then he estimates ∠CBD, the angle between the base of the cliff, himself, and Diana,
who is standing at point D. Diana estimates ∠CDB, the angle between the base of the cliff,
herself, and Brendan. Determine the height of the cliff to the nearest metre.
In ΔDBC,
∠BCD = 180° - 60° - 50°
∠BCD = 70°
BC
BD
=
sin D sin C
Þ
BC
60
=
sin 50 sin 70
BC = sin 50 (
60
) = 48.912...
sin 70
Example continued…
BC = sin 50 (
tan 76 =
AC
BC
Þ tan 76 =
60
) = 48.912...
sin 70
AC
48.912
Þ AC = 48.912 tan 76 = 196.177
The height of the cliff is 196 m.
PG. 161 – 164, #1-14, 16.
Independent Practice
Activity
Cut out the seven similar triangles from the handout, and use all
seven of them to form a single square. Show me once you succeed.
If the hypotenuse of the greatest triangle is 10 units long, what is
the area of the square?