Transcript Document 7209636

Solving Right Triangles
How do you solve right triangles?
M2 Unit 2: Day 6
Every right triangle has one right angle, two
acute angles, one hypotenuse, and two legs.
To SOLVE A RIGHT TRIANGLE means to
find all 6 parts.
To solve a right triangle you need…..
1 side length and 1 acute angle measure
-or2 side lengths
Given one acute angle and one side:
•To find the missing acute angle, use the
Triangle Sum Theorem.
•To find one missing side length, write an
equation using a trig function.
•To find the other side, use another trig
function or the Pythagorean Theorem
Solve the right triangle. Round decimal answers to
GUIDED PRACTICE
the nearest tenth.
Example 1
Find m∠ B by using the
Triangle Sum Theorem.
180o
=
90o
+
48o = m∠ B
42o
+ m∠ B
A
42o
70
48o
B
Approximate BC by using a tangent ratio.
C
Approximate AB by using a cosine ratio.
BC
o
tan 42 =
70
70
cos 42o =
AB
ANSWER
70 tan 42o = BC
AB cos 42o =
70
BC
70 0.9004
70
The angle measures are
= cos 42o 42o, 48o, and 90o. The
AB
63.0 ≈ BC
70
side lengths are 70 feet,
AB
0.7431 about 63.0 feet, and
about 94.2 feet.
94.2
AB
Solve a right triangle that has a 40o angle and a 20
GUIDED PRACTICE
inch hypotenuse.
Example 2
Find m∠ X by using the
X
Triangle Sum Theorem.
50o
180o = 90o + 40o + m∠ X
50o = m∠ X
20 in
Approximate YZ by using a sine ratio.
XY
sin 40o =
20
o
20 ● sin 40 = XY
20 ● 0.6428 ≈ XY
12.9 ≈ BC
Approximate AB by using a cosine ratio.
YZ
cos 40o =
20
o
20 ● cos 40 =
YZ
20 ● 0.7660 ≈ YZ
15.3 ≈ YZ
40o
Y
Z
ANSWER
The angle measures are 40o, 50o,
and 90o. The side lengths are 12.9
in., about 15.3 in., and 20 in.
Solve the right triangle. Round to
the nearest tenth.
Example 3
mQ  53
mR  90
mP  37°
PQ  30
PR  24.0
p
cos53 
30
p  18.1
q
30
q  24.0
sin53 
QR  18.1
If you know the sine, cosine, or tangent of an acute
angle measure, you can use the inverse
trigonometric functions to find the measure of the
angle.
Calculating Angle Measures from
Trigonometric Ratios
Example 4
Use your calculator to find each angle measure to the
nearest tenth of a degree.
A. cos-1(0.87)
cos-1(0.87)  29.5°
B. sin-1(0.85)
C. tan-1(0.71)
sin-1(0.85)  58.2°
tan-1(0.71)  35.4°
Inverse trig functions:
Ex: Use a calculator to approximate the
measure of the acute angle. Round to the
nearest tenth.
1. tan A = 0.5
mA  tan1(0.5) 
26.6°
2. sin A = 0.35
mA  sin1(0.35) 
20.5°
3. cos A = 0.64
mA  cos1(0.64) 
50.2°
EXAMPLE 2
Use an inverse sine and
an inverse cosine
Example 5
Let ∠ A and ∠ B be acute angles in a right triangle. Use a
calculator to approximate the measures of ∠ A and ∠ B to the
nearest tenth of a degree.
a.
sin A = 0.87
b.
SOLUTION
a.
m∠
A
= sin
b.
m∠
B
= cos
–1
–1
0.87 ≈ 60.5o
0.15≈ 81.4o
cos B = 0.15
Solving Right Triangles
Example 6
Find the unknown measures. Round
lengths to the nearest hundredth and
angle measures to the nearest degree.
Method 1: By the Pythagorean Theorem,
Method 2:
RT2 = RS2 + ST2
(5.7)2 = 52 + ST2
Since the acute angles of a right
triangle are complementary, mT 
90° – 29°  61°.
, so ST = 5.7 sinR.
Since the acute angles of a right
triangle are complementary,
mT  90° – 29°  61°.
Solve the right triangle. Round
decimals the nearest tenth.
Example 7
Use Pythagorean Theorem to find c…
c 2  22  33
c  3.6
Use an inverse trig function to
find a missing acute angle…
3
mA  tan ( )  56.3
2
1
Use Triangle Sum Theorem to
find the other acute angle…
mB  90  56.3  33.7
AB  3.6
BC  3
AC  2
mA  56.3°
mB  33.7°
mC  90
Example 8
PN 2  112  18 2
PN  21.9
11
mN  tan ( )  31.4
18
1
mP  90  31.4  58.6
Example 9
232  TU 2  72
TU  21.9
7
mS  cos ( )  72.3
23
1
mU  90  72.3  17.7
Solve the right triangle. Round decimals
to the nearest tenth.
mP  90  37  53
PQ
sin37 
22
PQ  13.2
QR
cos37 
22
QR  17.6
mT  90  24  66
TR cos 24  33
tan 24 
AT
33
TR  14.7 AT  36.1
 Homework:
 Pg
174 (#4-22 even)