PreCalculus AlgebraCynthia Young
Chapter 4.2 Right Triangle Trigonometry
Chapter 4 Section 2 Right Angle Trigonometry
Isosceles Right Triangle: a right triangle with two sides that
are congruent. 1) Find the measures of the base angles.
x x
h
yβ° yβ°
π¦β + π¦β + 90 = 180
2π¦β = 90 π¦ = 45β
x x
h
45β° 45β°
2) βXβ can be any number. To make it really easy, lets just
make x = 1.
h
45β° 45β°
1 1
3) Solve for βhβ.
222 cba 222 11 c
22 c
2c
45β° 45β°
1 1
2
βOne-One-Two-Rootβ
Use scale factors or proportions to solve for the lengths of sides
of similar 45-45-90 right triangles.
45
45
x1
12
45
45
Write a proportion (equation where
a fraction equals a fraction)
3 14
2=π₯
1
143x
3 2 7
2= π₯
π₯ = 3 7
Use scale factors or proportions to solve for the lengths of
sides of similar 45-45-90 right triangles.
45
45
x1
12
45
45
Write a proportion (equation where a fraction equals a fraction)
4 10
2=π₯
1x
4 2 5
2= π₯
π₯ = 4 5
4 10
60ΒΊ
2
60ΒΊ
30ΒΊ
2
2
1 1
30ΒΊ
By constructing angle bisector of the top angle of a 60-60-60
equilateral triangle, two triangles are formed.
Corresponding parts of congruent triangles are
congruent (CPCTC) (all remaining corresponding
pairs of angles and sides are congruent).
Are the two triangles congruent? Why? ASA
Length = 1 and length = 1
Bottom legs (of the right
triangles) are congruent so each
is Β½ the total of the original
triangleβs bottom length).
2x
1
30
60
We now have a 30-60-90 triangle.
Solve for βxβ.222 cba
π₯2 + 12 = 22
π₯2 = 4 β 1
π₯2 = 3π₯ = 3
βone-two-three-rootβ2
1
30
603
30-60-90 Right Triangle
60
30
1
Solve with a proportion
Write a proportion (equation where a fraction equals a
fraction)
60
30
X
y
6 2
2=π₯
1
2 β 3 β 2
2=π₯
1
6 2 2 3
3 2 = π₯
6 2
2=
π¦
3
2 β 3 β 2
2
3 2 =π¦
3
3 2 =π¦
3
β
β
β
3 6 = π¦β
β β
β 3 β3
1
Special Triangles
Solving Right Triangles
Solving Right Triangles given two sides
π πππ½ =πππ
βπ¦π=19.67 ππ
37.21 ππ
π πππ½ = 0.5286
sinβ1( π πππ½) = sinβ1(0.5286)
π½ = 37.9
πΌ = 180 β π½
πΌ = 148.1
πππ π½ =πππ
βπ¦π
πππ 37.9 =π
37.21 ππ
0.7891 =π
37.21 ππ
π = 29.4 ππ
Using Bearing
-Notice that the bearing (direction the plane is pointed) is relative to
north. We can assume that the position of the control tower is at the
location where the plane took off.
-We can draw a right triangle with leg lengths 5 and 12 miles since the
plane turned 90 degrees to the left between the initial bearing of 28 N and
the next bearing.
-We can calculate the bottom angle of the right triangle using the tangent
ratio.π = tanβ1
12
5= 67.4
To find the planeβs bearing from the control tower,
we must subtract the planeβs initial bearing from
theta. π½ = 67.4 β 28 π½ = 39.4We can describe the bearing as 39.4 degrees to
the west of north: (N 39.4 W) or using the true
direction it is 39.4 degrees to the left of north
(39.4 less than 360) or 320.6 βtrue.β
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