Transcript 13.1 Special Right Triangles

Textbook: Chapter 13
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Parts Of A Right Triangle
Hypotenuse
Acute
Angles
Leg
Right Angle
Leg
The Pythagorean Theorem
In a right triangle, the square of the length of the
hypotenuse is equal to the sum of the squares of
the lengths of the legs.
“No man is free who cannot control himself.”
― Pythagoras of Samos
Pythagorean Theorem
(leg)2 + (leg)2 = (hypotenuse)2 or a2 + b2 = c2
E1.) a = 3, b = 4 and c = ?
32 + 42 = 𝑐 2
𝑐=5
E2.) a = ? , b = 36 and c = 39
𝑎2 + 362 = 392
𝑎 = 15
E3.) a = 2 5, b = ? and c = 12
(2 5)2 +𝑏2 = 122
𝑏 = 2 31
P1.) a = 5, b = ? and c = 13
P2.) a = 7, b = 3 and c = ?
P3.) a = ?, b = 3 and c = 3 2
The sides of special right triangles (45-45-90 and 30-6090) have special relationships (ratios)
Given one side of a certain right triangle, we can use
these relationships (ratios) to unlock the other two
sides.
Why is this important if we already know the
Pythagorean Theorem?
We must know 2 sides of a right triangle to use the
Pythagorean Theorem.
45-45-90
If you are given a:
Leg (S)

The other leg is the same

Multiply by 2 to find the L (hypotenuse)
Hypotenuse (L)

Divide by 2 to find the S (legs)
b=5
c=5 2
a=
9 2
b=9
2
=9
a=2 6
b = 2 12 = 4 3
30-60-90
If you are given a:
Small Leg (S)


Multiply S by 3 to find M
Multiply S by 2 to find L
Medium Leg (M)


Divide M by 3 to find S
Multiply S by 2 to find L
Hypotenuse (L)


Divide L by 2 to find S
Multiply S by 3 to find M
*You need the S to unlock the M and L*
b=5 3
c = 10
a=8
b=8 3
a=
b=
10
=
3
20 3
3
10 3
3
Trigonometry
Trigonometry – measurement of triangles
Trigonometry Vocabulary
-based on the acute angle (𝜃) that is being used
From ∡𝐴
From ∡𝐵
Opposite
Hypotenuse
Adjacent
Adjacent
Hypotenuse
Opposite
Trigonometric Ratio – a ratio of the lengths of
two sides of a triangle.
Sine (sin), Cosine (cos) and Tangent (tan) are
the 3 basic trigonometric ratios
Sin
Cos
Tan
Solving Trigonometric Equations
E1) Variable on top (multiply both sides by the denominator)
𝑥
5
𝑥 = 5𝑐𝑜𝑠32°
𝑥 ≈ 4.24
𝑐𝑜𝑠32° =
E2) Variable on bottom (flip-flop the trig function and the variable)
13
𝑠𝑖𝑛42° =
𝑥
13
𝑥=
𝑠𝑖𝑛42°
𝑥 ≈ 19.43
E3) Variable attached to the trigonometric function (inverse both sides)
12
13
12
θ = tan−1
13
𝜃 ≈ 42.71°
𝑡𝑎𝑛𝜃 =
Solving Trigonometric Equations
P1) Variable on top (multiply both sides by the denominator)
sin 73 ° =
𝑥
6
P2) Variable on bottom (flip-flop the trig function and the variable)
tan 17 ° =
13
𝑥
P3) Variable attached to the trigonometric function (inverse both sides)
𝑐𝑜𝑠𝜃 =
5
17
5
65
4
5
3
5
4
3
25
5
=
65 13
60 12
=
65 13
25
5
=
60 12
H
O
𝑥
15
𝑥 = 15sin 29°
𝑥 ≈ 7.27
𝑠𝑖𝑛 29° =
O
13
𝑡𝑎𝑛 25° =
𝑥
13
𝑥=
tan 25°
𝑥 ≈ 27.88
𝑆
𝑜
𝑎
𝑜
𝐶
𝑇
ℎ
ℎ
𝑎
A
H
A
7
13
7
Θ = cos −1
13
𝜃 ≈ 57.4°
𝑐𝑜𝑠 𝜃 =
𝑆
𝑜
𝑎
𝑜
𝐶
𝑇
ℎ
ℎ
𝑎
Solving a right triangle means to find the measure
of three angles of the triangle and three sides of
the triangle. In other words all six parts.
You can solve a right triangle if you know:
(1) Two side lengths OR
(2) One side length and one angle measure
A= 41°
B= 49°
C= 90°
a= 7.8
b= 9
c=11.9
Step 1: Find a missing side using the given
information (Find c)
B,b,c
9
sin 49° =
𝑐
9
𝑐=
sin 49°
𝑐 ≈ 11.9
Step 2: Find the other side (Find a)
B,b,a
9
tan 49° =
𝑎
9
𝑎=
𝑡𝑎𝑛49°
𝑎 ≈ 7.8
A=
B=
C=
27°29′
62°31′
90°
a= 6
b=11.5
c= 13
Step 1: Find the missing Side
62 + 𝑏 2 = 132
36 + 𝑏 2 = 169
𝑏 2 = 133
𝑏 = 133
𝑏 ≈ 11.5
Step 2: Find a missing angle using the given
information (Find A)
A,a,c
6
sin 𝐴 =
13
−1 6
A = sin
13
𝐴 ≈ 27°29′
B,a,c
𝑎
10
𝑎 = 10 cos 63°
𝑎 ≈ 4.5 𝑓𝑒𝑒𝑡
cos 63° =
Angle of Elevation vs. Angle of Depression
Angle of Elevation = Angle of Depression
Angle of Elevation vs. Angle of Depression
You are standing on top of a building that is 50 ft. tall and you
see your buddy on the street. He is standing 30 ft. from the base
of the building. Find the angle of depression between you and
your buddy.
50
tan 𝑥 =
30
50
−1
𝑥 = tan
30
𝑥 ≈ 59°
The angle of elevation is 59° and
because the angle of elevation is the
same as the angle of depression, then
the angle of depression is also 59°.
𝑥
60
𝑥 = 60 tan 65°
𝑥 ≈ 128.7 𝑓𝑡.
tan 65° =
𝑥
65°
60 ft
Angle of Depression
200 ft
𝜃
175 ft
200
175
200
𝜃 = tan−1
175
𝜃 ≈ 48°49′
tan 𝜃 =