Geometry and Trigonometric Functions: Solutions
Here are some solutions to the Exercises to accompany the section Geometry
and Trigonometric Functions.
The Geometry of Right Triangles
- Given the indicated measurements in a triangle labeled
as follows:
where C = p/2, determine all the remaining
measurements. Note: This picture should only be used to
indicate the names of angles and their corresponding opposite sides; the
actual measurements in the picture will be different from those in the
Exercises.
- If we know that a = 5 in. and b
= 3 in., compute approximate values for c, A,
and B.
- Solution
- By the Pythagorean
Theorem, c2 = 52
+ 32 = 34, so c
» 5.83 in.
Using the formula for tan, we
have tan(A) = a/b
= 5/3, so that A = tan-1(5/3)
» 1.03 radians. Note:
While we could use the formula for sin, this would be less accurate,
since it would depend on an approximate value in its
calculation. Likewise, tan(B)
= b/a
= 3/5, so that B = tan-1(3/5)
» 0.54 radians. Again, we
could solve for this a different way, by the angle sum formula B
= p/2 - A
» p/2 - 1.03
» 0.54 radians, but since this
depends on an approximate answer for its computation, it will lead
to a less accurate answer; in this case, the discrepancy is out in
the fourth decimal place and so it not very noticeable.
- If we know that b
= 3 ft. and c = 7 ft., compute approximate values for a, A,
and B.
- Solution
- By the Pythagorean
Theorem, 72 = a2
+ 32, so a2
= 40 and a » 6.32
ft. Using the formula for cos,
we have cos(A) = b/c
= 3/7, so that A = cos-1(3/7)
» 1.13 radians.
Using the formula for sin, we
have sin(B) = b/c
= 3/7, so that B = sin-1(3/7)
» 0.44 radians.
- If we know that A = 0.8 radians
and b
= 3 m., compute approximate values for a, c,
and B.
- Solution
- By the angle sum formula B = p/2 -
A » p/2 -
0.8
» 0.77
radians. Using the formula
for cos, we have cos(A) =
cos(0.8) = b/c
= 3/c, so that c = 3/cos(0.8)
» 4.3 m.
Using the formula for tan gives
tan(A) = tan(0.8)
= a/b
= a/3,
so that a = 3tan(0.8)
» 3.09 m.
Back to Exercises.
- Solve for the unknown measurement in each of the
following pictures. Note: This pictures are not drawn to
scale; they simply intended to provide an applied context for the question.
- Find the height, h , (to the roofline) of the building:
.
- Solution
-
Using the formula for tan gives
tan(0.6) = a/b
= h/30,
so that h = 30tan(0.6)
» 20.5 ft.
- Find how high, h , the kite is off
the ground:
.
- Solution
-
Using the formula for sin, we
have sin(0.8) = h/50,
so that h = 50sin(0.8)
» 35 ft.
- Find the angle, A , the 6 ft.
ladder makes with the ground:
.
- Solution
-
Using the formula for cos, we
have cos(A) = 3/6
= 0.5, so that A = cos-1(0.5)
» 1.05 radians. In fact, we
recognize 0.5 as the exact value of cos at the angle p/3.
Back to Exercises.
Two General Formulas for Triangles
- Given the indicated measurements in a triangle labeled
as follows:
determine all the remaining measurements. Note: This picture
should only be used to indicate the names of angles and their
corresponding opposite sides; the actual measurements in the picture will be
different from those in the Exercises.
- If we know that a = 5 ft., b
= 3 ft., and c
= 7 ft., compute approximate values for C, A,
and B.
- Solution
- Using the Law of Cosines, 72
= 52 + 32
- 2·5·3cos(C),
so that C = cos-1((72 - 52
- 32)/(-2· 5·3))
= cos-1(-1/2) » 2.09
radians. Again, we can recognize -0.5 as the exact value of
cos at the angle 2p/3.
- If we know that a = 5 ft., b
= 3 ft., and c
= 9 ft., compute approximate values for C, A,
and B.
- Solution
- Again, using the Law of Cosines,
92 = 52 + 32
- 2· 5· 3cos(C),
so that C = cos-1((92 - 52
- 32)/(-2· 5· 3))
= cos-1(-47/30), which is undefined. Since the
shortest distance between two points is a straight line, it is not
possible to have a + b
< c, but 5 + 3
< 9. This all means that there cannot be a triangle
with such measurements.
- If we know that a = 80 mi., B
= 0.25 radians, and C = 0.65 radians, compute approximate
values for A, c, and b.
- Solution
- By the angle sum formula A = p -
B - C = p -
0.25 - 0.65 » 2.24.
Using the Law of Sines, b/a
= sin(B)/sin(A),
so that b = asin(B)/sin(A)
» 80sin(0.25)/sin(2.24)
» 25.24 mi.
Likewise, c/a = sin(C)/sin(A),
so that c = asin(C)/sin(A)
» 80sin(0.65)/sin(2.24)
» 61.73 mi.
-
- If we know that a = 15 m., A
= 0.4 radians, and c = 30 m., compute approximate values
for C, B, and b.
- Solution
- Using the Law of Sines, c/a = sin(C)/sin(A),
so that sin(C) = csin(A)/a
» 30sin(0.4)/15
» 0.78. We know that 0 < C
< p - A
» 2.74. The inverse
sin formula, C = 2kp
+ sin-1(0.78), (2k + 1)p - sin-1(1.04),
gives two valid solutions in this interval, when k = 0, C
» 0.90,
2.25. By the angle sum formula, this gives two different
values for B = p -
A - C » 1.85,
0.50. Using the Law of Sines
again, we can solve for the two possible values for b:
b/a = sin(B)/sin(A),
so that b = asin(B)/sin(A)
» 15 sin(1.85)/sin(0.4),
15 sin(0.50)/sin(0.4)
» 37 m.,
18.5. These measurements
correspond to the two triangles:
-
and 
.
-
Note: It is common to get two solutions, like
this, when you are given two sides and an angle, if the angle is
opposite to one of the sides; in High School geometry this situation
is described by the Side-Side-Angle Theorem.
-
Back to Exercises.
- Solve for the unknown measurement in each of the
following pictures. Note: This pictures are not drawn to
scale; they simply intended to provide an applied context for the question.
- Find the length, L, of the lake:
(top view).
- Solution
- Using the Law of Cosines, L2
= 302 + 202
- 2·30·20cos(0.7) » 382,
so that L » 19.5
ft.
- Find the distance, d, to the top of the
mountain:
(side view)
- Solution
- By the angle sum formula, the angle opposite the 500 m. side has
measure p - 0.4 - 2.5 » 0.24.
Using the Law of Sines, d/500
= sin(2.5)/sin(0.24),
so that d = 500sin(2.5)/sin(0.24)
» 1260 m.
- Use your answer to part b) to find the height, h, of the
mountain:
.
- Solution
-
Using the formula for sin,
sin(0.4) = h/1260,
so that h
= 1260sin(0.4) » 490 m.
Back to Exercises.
Go to Algebraic
Functions.
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