2 Dec 2015 9:50 - 11:20 am
Geometry Agenda
1) Review Using Sine, cosine, tangent to solve
problems
2) Review Pythagorean theorem and introduce
special right triangles
Objectives
Individual :
Students more responsible for learning correcting - reflecting - teaching - editing
Academic:
Geometric ratios, Pythagorean theorem and
Trigonometric ratios
For right triangles
sine
sin θ = opposite leg
hypotenuse
c
θ
For right triangles
sin θ = opposite leg
hypotenuse
cos θ = adjacent leg
hypotenuse
c
θ
cosθ
For right triangles
c
tanθ
sin θ = opposite leg
hypotenuse
cos θ = adjacent leg
hypotenuse
tan θ = opposite leg
θ
adjacent leg
For right triangles
sine
c
cosθ
tanθ
sin θ = opposite leg
hypotenuse SOH
cos θ = adjacent leg
hypotenuse
tan θ = opposite leg
θ
adjacent leg
For right triangles
sine
c
cosθ
tanθ
sin θ = opposite leg
hypotenuse SOH
cos θ = adjacent leg
hypotenuse
tan θ = opposite leg
θ
adjacent leg
CAH
For right triangles
sine
c
cosθ
tanθ
sin θ = opposite leg
hypotenuse SOH
cos θ = adjacent leg
hypotenuse
CAH
tan θ = opposite leg
θ
adjacent leg
TOA
SOH CAH TOA
Review
1. Solve for the side x. Round all final answers to
2 decimal places.
B
12cm
38°
C
x
A
Review
1. Solve for the side x. Round all final answers to
2 decimal places.
side opposite angle
B
side adjacent to angle
x
12cm
38°
C
A
Review
1. Solve for the side x. Round all final answers to
2 decimal places.
side opposite angle
B
side adjacent to angle
x
12cm
38°
C
A
x cm = tan 38°
12cm
Review
1. Solve for the side x. Round all final answers to
2 decimal places.
side opposite angle
B
side adjacent to angle
x
12cm
38°
C
A
x
= tan 38°
12cm
x = 12cm(tan38°)
Use calculator
Review
1. Solve for the side x. Round all final answers to
2 decimal places.
side opposite angle
B
side adjacent to angle
x
12cm
38°
C
A
x
= tan 38°
12cm
x = 12cm(tan38°)
x = 9.375427cm
Review
1. Solve for the side x. Round all final answers to
2 decimal places.
side opposite angle
B
side adjacent to angle
x
12cm
38°
C
A
x
= tan 38°
12cm
x = 12cm(tan38°)
x = 9.375427cm
x approx = 9.38cm
2. Find the missing angle. Round to nearest
tenths.
32cm = hypotenuse
θ°
32cm
c
25cm
25cm = side
adjacent to angle θ°
2. Find the missing angle. Round to nearest
tenths.
32cm = hypotenuse
θ°
32cm
c
25cm
25cm = side
adjacent to angle θ°
cosθ = 25cm
32cm
2. Find the missing angle. Round to nearest
tenths.
32cm = hypotenuse
θ°
32cm
c
25cm
25cm = side
adjacent to angle θ°
cosθ = 25cm
32cm
cosθ = 0.78125
2. Find the missing angle. Round to nearest
tenths.
32cm = hypotenuse
θ°
32cm
c
25cm
cosθ = 0.78125
θ = cos-1(0.78125)
calculator, inverse
cos0.78125
2. Find the missing angle. Round to nearest
tenths.
32cm = hypotenuse
θ°
32cm
c
θ
= 38.6248°
25cm
cosθ = 0.78125
θ = cos-1(0.78125)
calculator, inverse
cos0.78125
2. Find the missing angle. Round to nearest
tenths.
32cm = hypotenuse
θ°
32cm
c
θ
25cm
cosθ = 0.78125
θ = cos-1(0.78125)
calculator, inverse
cos0.78125
= 38.6248° approx. = 38.6°
3. How far from the base of a building
is the bottom of a 30ft ladder that
makes a 75°angle with the ground?
3. How far from the base of a building
is the bottom of a 30ft ladder that
makes a 75°angle with the ground?
30ft
75°
x
3. How far from the base of a building
is the bottom of a 30ft ladder that
makes a 75°angle with the ground?
cos 75° = side adjacent to angle
hypotenuse
30ft
75°
x
3. How far from the base of a building
is the bottom of a 30ft ladder that
makes a 75°angle with the ground?
cos 75° = side adjacent to angle
hypotenuse
30ft
75°
x
cos 75° = x
30feet
30feet cos75° = x
calculator
3. How far from the base of a building
is the bottom of a 30ft ladder that
makes a 75°angle with the ground?
cos 75° = side adjacent to angle
hypotenuse
30ft
75°
x
cos 75° = x
30feet
30feet cos75° = x
7.76ft =x
x = 7.76 ft
4. A little boy flies his kite. The string
forms an angle of elevation of 37
degrees and from where he stands to
directly under the kite is 45 feet. How
long is the kite string to the nearest
foot? Diagram provided on next slide.
How long is the kite string (hypotenuse) to the
nearest foot (nearest whole number)?
x
37°
45feet
5. A 3.4ft guy wire is attached to a tree 3ft from
the ground. What is the angle formed between
the wire and the ground, to the nearest degree?
3.4feet
θ
3ft
2nd look at
Homework that was due Weds Nov 18
8-4 Trigonometry
(pages 460 - 461)
24, 25
29, 30, 31, and 34
44, 46, 47, and 48.
Also now do 50 and 51 (with diagrams).
Right Triangle Trigonometry 3
Review and Practice
1. Find x.
a)
b)
c)
Use Pythagorean theorem each time.
Label vertices of triangles and sides.
Special Right Triangles
2. Draw a square.
Label the sides x.
Label the 4 right angles.
Draw a diagonal.
What postulate confirms that we have now got
2 congruent triangles?
Special right triangles cont.
x
x
x
x
Special right triangles cont.
x
x
h
By Pythagorean
theorem
h2 = x 2 + x 2
x h2 = 2x2
h = √2√x2
h = x√2
x
Special right triangles 45°-45°-90°.
45°
n
n√2
n
45°
We have shown
that for any 45°45°
90° triangle, its
hypotenuse is √2
times the length of
its leg.
Ratio of sides 1:1:
√2
2. Find x.
45°
x
3√2
x2 + x2 = (3√2)2
2x2 = 32√2√2
x2 = 9(2)
2
2. Find x.
45°
3√2
x
x = √9
x=3
45°
x
2. Find x. OR
45°
x
x
For any 45°45°90°
triangle,
hypotenuse is √2
times the length of
its leg.
3√2 Ratio of sides
1:1:√2
x:x:x√2
here x:x:3√2
45°
so x = 3
3. Find x and y.
45°
14
x
45°
y
Special Right Triangles 30°-60°-90°
Draw an equilateral triangle with sides 2x. Draw
a line that bisects one of the vertex angles and
a base (an altitude). Prove that what you’ve
drawn are 2 congruent 30°-60°-90° triangles.
Use the Pythagorean theorem to find a and so
find the ratio of the sides in a 30°-60°-90°
triangle.