Download 183 7–4 Solution of Right Triangles Pythagorean Theorem Sum of

Section 7–4
◆
183
Solution of Right Triangles
Without using tables or a calculator, write the sin, cos, and tan of angle A. Leave your answer
in fractional form.
3
12
12
7. sin A 8. cot A 9. cos A 13
5
5
Evaluate the following, giving your answer in decimal degrees to three significant digits.
10. arcsin 0.635
13. cot
1 1.17
7–4
12. tan
1 2.85
15. arccsc 4.26
11. arcsec 3.86
14. cos
1 0.229
Solution of Right Triangles
Right Triangles
The right triangle (Fig. 7–10) was introduced in Sec. 6–2, where we also introduced
the Pythagorean theorem (Eq. 145) and the equation for the sum of the interior angles
(Eq. 139).
Since the angle C is always 90 for right triangles, the equation for the sum of the
interior angles can be re-written as
A B 90 180
A
c
b
C
or
A B 90
This plus our trigonometric relations for sin, cos, and tan become our tools for solving
any right triangle.
Pythagorean
Theorem
c2 a2 b2
145
Sum of the
Angles
A B 90
139
Trigonometric
Functions
side opposite to sin hypotenuse
146
side adjacent to cos hypotenuse
147
side opposite to tan side adjacent to 148
Solving Right Triangles When One Side and One Angle Are Known
To solve a triangle means to find all missing sides and angles (although in most practical problems we need find only one missing side or angle). We can solve any right triangle if we know
one side and either another side or one angle.
To solve a right triangle when one side and one angle are known:
1. Make a sketch.
2. Find the missing angle by using Eq. 139.
a
B
FIGURE 7–10 A right triangle. We will usually label a right
triangle as shown here. We label
the angles with capital letters
A, B, and C, with C always the
right angle. We label the sides
with lowercase letters a, b, and
c, with side a opposite angle
A, side b opposite angle B, and
side c (the hypotenuse) opposite
angle C (the right angle).
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Chapter 7
◆
Right Triangles and Vectors
3. Relate the known side to one of the missing sides by one of the trigonometric ratios. Solve
for the missing side.
4. Repeat Step 3 to find the second missing side.
5. Check your work with the Pythagorean theorem.
◆◆◆
Example 15: Solve right triangle ABC if A 18.6 and c 135.
Solution:
1. We make a sketch as shown in Figure 7–11.
2. Then, by Eq. 139,
B
35
1
c =
18.6°
A
B 90 A 90 18.6 71.4
a
b
FIGURE 7–11
Realize that either of the two
legs can be called opposite or
adjacent, depending on which
angle we are referring them to.
Here b is adjacent to angle A but
opposite to angle B.
3. Let us now find side a. We must use one of the trigonometric ratios. But how
do we know which one to use? And further, for which of the two angles, A or B,
C
should we write the trig ratio?
It is simple. First, always work with the given angle, because if you made a mistake in
finding angle B and then used it to find the sides, they would be wrong also. Then, to decide
which trigonometric ratio to use, we note that side a is opposite to angle A and that the given
side is the hypotenuse. Thus, our trig function must be one that relates the opposite side to
the hypotenuse. Our obvious choice is Eq. 146.
opposite side
sin A hypotenuse
Substituting the given values, we obtain
a
sin 18.6 135
Solving for a yields
a 135 sin 18.6
135(0.3190) 43.1
4. We now find side b. Note that side b is adjacent to angle A. We therefore use Eq. 147.
b
cos 18.6 135
As a rough check of any triangle,
see if the longest side is
opposite the largest angle and if
the shortest side is opposite the
smallest angle. Also check that
the hypotenuse is greater than
each leg but less than their sum.
So
b 135 cos 18.6
135(0.9478) 128
We have thus found all missing parts of the triangle.
5. For a check, we see if the three sides will satisfy the Pythagorean theorem.
Check:
(43.1)2 (128)2 (135)2
?
18 242 18 225
?
A
Since we are working to three significant digits, this is close enough for a check.
c
b
◆◆◆
B
55.2°
a = 207
FIGURE 7–12
C
Example 16: In right triangle ABC, angle B 55.2 and a 207. Solve the triangle.
Solution:
1. We make a sketch as shown in Fig. 7–12.
2. By Eq. 139,
A 90 55.2 34.8
◆◆◆
Section 7–4
◆
185
Solution of Right Triangles
3. By Eq. 147,
207
cos 55.2 c
207
207
c 363
cos 55.2 0.5707
4. Then, by Eq. 148,
b
tan 55.2 207
b 207 tan 55.2 207(1.439) 298
5. Checking with the Pythagorean theorem, we have
(363)2 (207)2 (298)2
?
131 769 131 653 (checks to within three significant digits)
?
Tip
◆◆◆
Whenever possible, use the given information for each computation,
rather than some quantity previously calculated. This way, any errors
in the early computation will not be carried along.
This is good advice for performing
any computation, not just for
solving right triangles.
Solving Right Triangles When Two Sides Are Known
1. Draw a diagram of the triangle.
2. Write the trigonometric ratio that relates one of the angles to the two given sides. Solve for
the angle.
3. Subtract the angle just found from 90 to get the second angle.
4. Find the missing side by the Pythagorean theorem.
5. Check the computed side and angles with trigonometric ratios, as shown in the following
example.
◆◆◆
Example 17: Solve right triangle ABC if a 1.48 and b 2.25.
Solution:
1. We sketch the triangle as shown in Fig. 7–13.
2. To find angle A, we note that the 1.48 side is opposite angle A and that the 2.25 side is
adjacent to angle A. The trig ratio relating opposite and adjacent is the tangent (Eq. 148).
1.48
tan A 0.6578
2.25
from which
A
c
A 33.3
b = 2.25
3. Solving for angle B, we have
B 90 A 90 33.3 56.7
4. We find side c by the Pythagorean theorem.
c2 (1.48)2 (2.25)2 7.253
c 2.69
Check:
? 1.48
sin 33.3 2.69
0.549 ⬇ 0.550
(checks)
B
a = 1.48
FIGURE 7–13
C
186
Chapter 7
◆
Right Triangles and Vectors
and
? 2.25
sin 56.7 2.69
0.836 0.836
(checks)
(Note that we could just as well have used another trigonometric function, such as the cosine,
◆◆◆
for our check.)
Cofunctions
The sine of angle A in Fig. 7–10 is
a
sin A c
But a/c is also the cosine of the complementary angle B. Thus we can write
sin A cos B
Here sin A and cos B are called cofunctions. Similarly, cos A sin B. These cofunctions and
others in the same right triangle are given in the following boxes:
Cofunctions
Where
A ⴙ B ⴝ 90ⴗ
or
P
A ⴙ B ⴝ rad
2
sin A cos B
154a
cos A sin B
154b
tan A cot B
154c
cot A tan B
154d
sec A csc B
154e
csc A sec B
154f
In general, a trigonometric function of an acute angle is equal to the corresponding cofunction of the complementary angle.
Example 18: If the cosine of the acute angle A in a right triangle ABC is equal to 0.725,
find the sine of the other acute angle, B.
◆◆◆
Solution: By Eq. 154b,
sin B cos A 0.725
Exercise 4
◆
◆◆◆
Solution of Right Triangles
Right Triangles with One Side and One Angle Given
Sketch each right triangle and find all of the missing parts. Assume the triangles to be labelled
as in Fig. 7–10. Work to three significant digits.
1. a 155
3. a 1.74
A 42.9
B 31.9
2. b 82.6
4. b 7.74
B 61.4
A 22.5
Section 7–5
◆
5. a 284
7. b 9.26
9. b 82.4
187
Applications of the Right Triangle
A 1.13 rad
B 55.2
A 31.4
6. b 73.2
8. a 1.73
10. a 18.3
B 0.655 rad
A 39.3
B 44.1
Right Triangles with Two Sides Given
Sketch each right triangle and find all missing parts. Work to three significant digits and express
the angles in decimal degrees.
11.
13.
15.
17.
19.
a 382
b 3.97
a 27.4
a 41.3
b 228
b 274
c 4.86
c 37.5
c 63.7
c 473
12.
14.
16.
18.
20.
a 3.88
a 63.9
b 746
a 4.82
a 274
c 5.37
b 84.3
c 957
b 3.28
c 429
Cofunctions
Express as a function of the complementary angle.
21. sin 38
22. cos 73
24. sec 85.6
25. cot 63.2
28. cos 0.153 rad
27. tan 3514
7–5
23. tan 19
26. csc 82.7
29. sin 0.475 rad
Applications of the Right Triangle
There is, of course, a huge number of applications for the right triangle, a few of which are
given in the following examples and exercises. A typical application is finding a distance that
cannot be measured directly, as shown in the following example.
B
◆◆◆
Example 19: To find the height of a flagpole (Fig. 7–14), a person measures 12.0 m
from the base of the pole and then measures an angle of 40.8 from a point 2.00 m above
the ground to the top of the pole. Find the height of the flagpole.
x
A
Estimate: If angle A were 45, then BC would be the same length as AC, or 12 m. But
our angle is a bit less than 45, so we expect BC to be less than 12 m, say, 10 m. Thus
our guess for the entire height is about 12 m.
Solution: In right triangle ABC, BC is opposite the known angle, and AC is adjacent.
Using the tangent, we get
x
tan 40.8 12.0
12.0 m
FIGURE 7–14
where x is the height of the pole above the observer. Then
x 12.0 tan 40.8 10.4 m
Adding 2.0 m, we find that the total pole height is 12.4 m, measured from the ground.
40.8°
2.00 m
◆◆◆
◆◆◆
Example 20: From a plane at an altitude of 1050 m, the pilot observes the angle of depression of a lake to be 18.6. How far is the lake from a point on the ground directly beneath the
plane?
Solution: We first note that an angle of depression, like an angle of elevation, is defined as the
angle between a line of sight and the horizontal, as shown in Fig. 7–15. We then make a sketch
for our problem (Fig. 7–16). Since the ground and the horizontal line drawn through the plane
C