Solving Right Triangles M2 Unit 2: Day 6 How do you solve right triangles?
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Solving Right Triangles
M2 Unit 2: Day 6
How do you solve right triangles?
To solve a right triangle you need…..
1 side length and 1 acute angle measure-or-
2 side lengths
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.
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
GUIDED PRACTICE
Example 1
Find m∠ B by using the Triangle Sum Theorem.
180o = 90o + 42o + m∠ B
48o = m∠ B
A
C B48o
7042o
Solve the right triangle. Round decimal answers to the nearest tenth.
Approximate BC by using a tangent ratio.
tan 42o =BC 70
70 tan 42o = BC
70 0.9004 BC
63.0 ≈ BC
Approximate AB by using a cosine ratio.
cos 42o =70 AB
AB cos 42o = 70
AB 70 cos 42o=
AB 70 0.7431
AB 94.2
ANSWER
The angle measures are 42o, 48o, and 90o. The side lengths are 70 feet, about 63.0 feet, and about 94.2 feet.
GUIDED PRACTICESolve a right triangle that has a 40o angle and a 20
inch hypotenuse.
Example 2
Find m∠ X by using the
Triangle Sum Theorem.
180o = 90o + 40o + m∠ X50o = m∠ X
X
YZ
Approximate YZ by using a sine ratio.
sin 40o =XY
2020 ● sin 40o = XY
20 ● 0.6428 ≈
XY
12.9 ≈ BC
Approximate AB by using a cosine ratio.
cos 40o =YZ 20
20 ● cos 40o = YZ20 ● 0.7660 ≈ YZ
15.3 ≈ YZ
The angle measures are 40o, 50o, and 90o. The side lengths are 12.9 in., about 15.3 in., and 20 in.
ANSWER
40o
20 in50o
Solve the right triangle. Round to the nearest tenth.
53
90
30
m Q
m R
m P
PQ
PR
QR
cos53
30
p sin5330
q
37°
18.1
24.0
Example 3
18.1p 24.0q
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
Use your calculator to find each angle measure to the nearest tenth of a degree.
A. cos-1(0.87) B. sin-1(0.85) C. tan-1(0.71)
cos-1(0.87) 29.5° sin-1(0.85) 58.2° tan-1(0.71) 35.4°
Example 4
Inverse trig functions:
Ex: Use a calculator to approximate the measure of the acute angle. Round to the
nearest tenth.
26.6° 20.5° 50.2°
1tan (0.5)m A 1sin (0.35)m A 1cos (0.64)m A
1. tan A = 0.5 2. sin A = 0.35 3. cos A = 0.64
EXAMPLE 2 Use an inverse sine and an inverse cosine
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. cos B = 0.15
SOLUTION
a. m ∠ A = sin –1 0.87 ≈ 60.5o
b. m ∠ B = cos –1 0.15≈ 81.4o
Example 5
Solving Right Triangles
Find the unknown measures. Round lengths to the nearest hundredth and angle measures to the nearest degree.
Method 1: By the Pythagorean Theorem,
Since the acute angles of a right triangle are complementary, mT 90° – 29° 61°.
RT2 = RS2 + ST2
(5.7)2 = 52 + ST2
Method 2:
Since the acute angles of a right triangle are complementary, mT 90° – 29° 61°.
, so ST = 5.7 sinR.
Example 6
Solve the right triangle. Round decimals the nearest tenth.
3
2
90
AB
BC
AC
m A
m B
m C
Use Pythagorean Theorem to find c…
2 2 32 3
3.6
c
c
3.6Use an inverse trig function to find a missing acute angle…
1 3tan ( ) 56.3
2m A
Use Triangle Sum Theorem to find the other acute angle…
90 56.3 33.7m B
56.3°
33.7°
Example 7
2 2 211 18
21.9
PN
PN
1 11tan ( ) 31.4
18m N
90 31.4 58.6m P
Example 8
2 2 223 7
21.9
TU
TU
1 7cos ( ) 72.3
23m S
90 72.3 17.7m U
Example 9
Solve the right triangle. Round decimals to the nearest tenth.
90 37 53m P
sin3722
13.2
PQ
PQ
cos3722
17.6
QR
QR
90 24 66m T
tan243314.7
TR
TR
33cos24
36.1AT
AT
Homework:Pg 174 (#4-22 even)
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