Download Section 6: Solving Right Triangles

6. SOLVING RIGHT TRIANGLES
In the right triangle ACB shown in Figure 6.1, the angles are denoted by α at vertex A,
β at vertex B, and γ at vertex C. The lengths of the sides opposite the angles α , β ,
and γ are denoted by a, b, and c. Note that angles α and β are complementary acute
angles, that angle γ is a right angle, and that c is the hypotenuse of right triangle ACB.
Therefore,
sin α = cos β =
cos α = sin β =
tan α = cot β =
a
c
b
c
a
b
csc α = sec β =
sec α = csc β =
cot α = tan β =
•B
c
a
c
b
b
a
β
c
a
•
α
γ
b
A
•
C
Figure 6.1
If the lengths of two sides of a right triangle are given, or if one side and an acute angle
are given, then these formulas san be used to solve for the remaining angles and sides of
the triangle. The procedure is called solving the right triangle.
Example 6.1 ---------------------------- -----------------------------------------------------------Solve the right triangle ACB labeled as in Figure 6.2 if β = 30 o and a = 24.
•B
We must find b, c¸ and α . Because α and β are complementary,
α = 90 o − β = 90 o − 30 o = 60 o .
c 30
Figure 6.2
o
24
•
A
α
γ
b
•
C
To find b, we use the fact that tan β =
b
,
a
so b = a tan β .
 3
 =8 3 .
Therefore, b = a tan β = 24 tan 30 o = 24 
 3 


To find c, we use the fact that sec β =
Therefore,
c
a
, so c = a sec β .
2 3
 = 16
c = a sec β = 24 sec 30 o = 24 
 3 


Of course, we could have instead used the Pythagorean theorem to find c =
3 .
a2 + b2 .
_______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
Unless the special angles 30 o , 45 o , or 60 o are involved, it is necessary to use a calculator
or a table of trigonometric functions to solve a right triangle. You should always keep in
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mind that the solutions obtained in these ways are approximations. In this section,
whenever such approximations are involved, we will round off all angles to the nearest
hundredth of a degree, and all side lengths to four significant digits.
Example 6.2 ---------------------------- -----------------------------------------------------------In each case, assume that right triangle ACB is labeled as in Figure 6.1. Then solve the
triangle and sketch it.
( i ) b = 31, α = 43.33 o .
We must find a, c, and β . Because α and β are
complementary,
β = 90 o − α = 90 o − 43.33 o = 46.67 o .
a
To find a, we notice that tan α = , so a = b tan α .
b
Using a calculator, we find that
tan α = tan 43.33 o = 0.943341386.
Therefore, rounded off to four significant digits,
a = b tan α = 31 tan 43.33 o = 29.24 .
b
b
, so c =
.
c
cos α
Therefore, rounded off to four significant digits,
31
b
31
so c =
=
=
= 42.62 .
o
cos α
0.727413564
cos 43.33
To find c, we notice that cos α =
•B
β = 46.67 o
c=42.62
a=29.24
γ
α = 43.33o
•
•
C
b=31
A
Figure 6.3
Once again, we could have instead used the Pythagorean theorem to solve for c:
c =
a2 + b2 =
(29.24)2 + 312
= 42.61.
The discrepancy in the last decimal place was caused by using the rounded off value for a. The
result c = 42.62 is actually more accurate. The triangle is shown in figure 6.3.
( ii ) a = 4, b = 3 .
Figure 6.4
•B
By the Pythagorean theorem:
β = 36.87 o
a2 + b2 =
16 + 9 = 5.
4
a
Now we have sin α = = = 0.8 .
c
5
c =
c=5
α = arcsin 0.8 = 53.13 o .
It follows that
a=4
β = 90 o − α = 90 o − 53.13 o = 38.87 o .
The triangle is shown in Figure 5.4.
α = 53.13o
•
A
b=3
γ
•
C
_______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
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Example 6.3 ---------------------------- -----------------------------------------------------------Consider a church with a steeple, as shown in the Figure 6.5. The problem is to calculate
the height of the steeple while standing on the ground. To find the height, mark a point B
on the ground directly below the steeple and another point A that is 200 feet away on the
ground. At A , measure the angles of elevation α and β to the top and bottom of the
steeple. This is all the information you need.
Figure 6.5
Let x be the height of the steeple and y be the distance from the ground to the bottom of the steeple.
Suppose that α = 35 o and β = 26 o .
x+ y
y
and
tan 26 o =
.
Then tan 35 =
200
200
Solving for y in the second equation and substituting into the first
yields:
x + 200 tan 26 o
.
tan 35 o =
200
•
o
x
•
y
α = 35o
So x = 200 tan 35 o - 200 tan 26 o =200 (.7002-.4877) = 42.5 feet.
•
A
β = 26 o
•
B
200 ft
Section 6 Problems--------------- ------- -----------------------------------------------------------In problems 1-3, assume that the right triangle is
labeled as in the figure shown. In each case, solve
the triangle and sketch it. When approximations are
involved, round off angles to the nearest hundredth
of a degree and side lengths to four significant
digits.
1. a = 10, c =10 2
2. c =100, β = 41o
3. b = 91, β = 30 o
43
•B
β
c
a
•
A
α
γ
b
•
C
4. Find the area of the shaded region of the right
triangle ABC shown in the figure.
5. The house in the figure is 24 ft. wide.
The ridge is 9 ft higher than the side
walls, and the rafters project 1.5 ft
beyond the sides of the house. How
long are the rafters?
6. Find the distance AB between the centers of
the two rollers in the figure.
7. A highway cuts a corner from a parcel of land.
Find the number of acres in the triangular lot
ABC. (Note: 1 acre = 43,560 ft 2 ).
(Hint: First use the Law of Cosines, page 54, to find the
the measure of one angle in ∆ ABC.)
Hin
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8. A beam AB is supported by two crossed
beams. Find the distance x.
9. From the top of a lighthouse 120 feet above sea level, the angle of depression (the
downward angle from the horizontal) to a boat adrift ion the sea is 9.4 o . How far from
the foot of the lighthouse is the boat?
10. From a window in an office building, I am looking at a television tower that is 600
meters away (horizontally). The angle of elevation of the top of the tower is 19.6 o
and the angle of depression of the base of the tower is 21.3 o . How tall is the tower?
11. Find x in the figure.
12. The vertical distance from the first to the second floor of a certain department store is
28 feet. The escalator, which has a horizontal reach of 86 feet, takes 25 seconds to
carry a person between floors. How fast does the escalator travel?
13. Consider two circles both of radius r and with the center of one lying on the rim of
the other. Find the area of the common part of the two circles.
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14. The Great Pyramid is about 480 feet
high and its square base measures
760 feet on a side. Find the angle of
elevation, β , of one of its edges.
15. Find the angle between a principal diagonal and a face of a cube.
16. A regular hexagon (6 equal sides) is inscribed ia a circle of radius 4. Find the
perimeter, P , and the area, A, of this hexagon.
17. Find the area of the regular 6-pointed Star of
David that is inscribed in a circle of radius 1.
18. Find the area of the regular 5-pointed star (the
pentagram) that is inscribed in a circle of radius 1.
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