Trigonometry/ProOneSoftware-Trigonometry-Win31.iso/trig/chapter1.5r

Text File | 1995-04-09 | 10KB | 344 lines

 342
à 1.5ïSolving Oblique Triangles
ä Please find the indicated part of the given oblique
êëtriangle.
âêêê Find side "a" in the given figure.

êêêêêïaêè 42 ft
#êêêêë ───────è=è───────
êêêêë sin 40°êsin 110°
êêêêêïaè≈è28.73 ft
@fig1501.bmp,15,118

éSïIn this section, we will solve oblique triangles, which are
triangles with no right angles.ïYou can always solve an oblique tri-
angle if you know two angles and a side, two sides and the included an-
ge, or three sides.ïThis assumes for example, that three sides can form
a triangle.ïIn addition, a fourth case, where you know one angle and two
sides with one side opposite the given angle, will be considered.
è To solve an oblique triangle when two angles and a side are known,
we will need the Law of Sines.ïIn the figure, using triangle ACD,
êêê sin Aï=ïh/b or hï=ïb∙sin A.ïAlso, using tri-
@fig1502.bmp,15,150
êêê angle BCD, sin Bï=ïh/a or hï=ïa∙sin B.ïThe
êêê two expressions for "h" are set equal to each
êêêïother.ê b∙sin Aè=èa∙sin B
êêêêêë aêèb
#êêêêêè─────è=è─────
êêêêêèsin Aêsin B

êêê Using two different angles, a third term can
êêê be added to this equation to give us the
êêê "Law of Sines".ïRemember, you can use this
êêê equation to solve an oblique triangle
êêê whenever you know two angles and a side.
êêêïaêè bêè c
#êêê─────è=è─────è=è─────
êêêsin Aêsin Bêsin C
è In the example, two angles and a side are known.
êêè a/(sin 40°)è=è(42 ft)/(sin 110°)
êêêêaè≈è28.73 ft
Thus, the unknown side is 28.73 ft.ïIf you wanted to find the other
side, you could apply the Law of Sines using the third angle.
 1
êêêë Find side "c" in the given triangle.

êêêèA)ï10.6 ftêê B)ï9.63 ft

êêêèC)ï12.28 ftêêD)ïå of ç
@fig1503.bmp,25,118
üïIn this triangle, we know two angles and a side, so we can use
the Law of Sines to find unknown parts.
êêêë cêë 12 ft
#êêêï─────────è =è───────
êêêïsin 27.4°ê sin 35°

êêêêècè≈è9.63 ft
Ç B
 2
êêêë Find side "c" in the given triangle.

êêêèA)ï4.32 miêê B)ï3.79 mi

êêêèC)ï5.69 miêê D)ïå of ç
@fig1504.bmp,15,118
üïIn this figure we know two angles and a side, but it will be
necessary to find the third angle in order to use the Law of Sines to
find side "c".ïSince the sum of the interior angles of any triangle
must equal 180°, angle C is seen to be 111.3°.ïNow you can find "c".
êêêë cêë4.32 mi
#êêêï─────────è =è───────
êêêïsin 111.3°êsin 45°
êêêêècè≈è5.69 mi
Ç C
 3
êêêèFind the distance between the two house tops.

êêêèA)ï94.08 ftêêB)ï74.2 ft

êêêèC)ï106.3 ftêêD)ïå of ç
@fig1505.bmp,15,118
üïThe distance between the two house tops is inaccessible, but
you can measure angle A, angle B, and side AB.ïSince we know two an-
gles and a side, we can use the Law of Sines.
êêêë bêë20 ft
#êêêï─────────è =è─────────
êêêïsin 111.6°êsin 11.4°
êêêê bè≈è94.08 ft
Ç A
 4
êêêèFind the distance, AB, across the river.

êêêèA)ï88.4 ydêê B)ï77.88 yd

êêêèC)ï106.2 ydêêD)ïå of ç
@fig1506.bmp,25,118
ü

êêêë cêè 100 yd
#êêêï───────è =è─────────
êêêïsin 37°ê sin 50.6°

êêêê cè≈è77.88 ft
Ç B
 5ïA ship at point B is 10.2 miles due east of another boat at
point A.ïFrom the boat at point A, the bearing to a lighthouse is
N 52° E, and the bearing from ship B to the same lighthouse is N 17° W.
Find the distance from the boat at point A to the lighthouse.
êêè A)ï16.4 miêê B)ï10.45 mi

êêè C)ï3.56 miêê D)ïå of ç
üêêïFrom the figure, you can see that the angle at A
êêê is 38°, the angle at B is 73°, and the angle at C
êêê 69°.
êêêêë bêè 10.2 mi
#êêêêè───────è =è───────
êêêêèsin 73°ê sin 69°
@fig1507.bmp,100,620
êêêêêbè≈è10.45 mi
Ç B
ä Please find the indicated part of the given triangle.
âêêëFind side "c" in the figure.

#êêêëcìï=ï6ì + 8ì - 2∙6∙8∙cos 22°

êêêêë cï≈ï3.315

@fig1508.bmp,15,118
éSïIn order to solve a triangle when either two sides and the in-
@fig1509.bmp,15,118
cluded angle are known or three sides are known, you will need the Law
of Cosines.
êêêëIn the figure, sin C = y/b or y = b∙sin C.
êêêïAlso cos C = x/b or x = b∙cos C.ïApply the
êêêïPythagorean Theorem to triangle ABD.
#êêêêë cìï=ïyì + (a - x)ì
#êêêëcìï=ï(b∙sin C)ì + (a - b∙cos C)ì
êêêïThis equation simplifies to the following:
#êêêê cìï=ïaì + bì - 2∙a∙b∙cos C.

êêêïIf we completed the same steps, but with
êêêïdifferent angles, we could derive the
êêêïfollowing equations.
#êêêè aìï=ïbì + cì - 2∙b∙c∙cos A
#êêêè bìï=ïaì + cì - 2∙a∙c∙cos B
These three equations are referred to as the Law of Cosines, which can
be used to solve a triangle given two sides and the included angle or
when given three sides.ïOnce you use the Law of Cosines to find one
additional part, it is generally easier to use the Law of Sines to find
the remaining unknowns.
è In the example, two sides and the included angle are known.ïThese
#values are substituted into the equation, cì = aì + bì - 2∙a∙b∙cos C, to
find that c = 3.315.
 6
êêêêFind side "a" in the figure.

êêêè A)ï9.83 inêêB)ï14.3 in

êêêè C)ï12.58 inêë D)ïå of ç
@fig1510.bmp,15,118
ü
#êë aìï=ï(22.3)ì + (29.6)ì - 2∙(22.3)∙(29.6)∙cos 23°

#êêêêaìï≈ï158.2363

êêêêïaï≈ï12.58 in
Ç C
 7
êêêêFind angle A in the figure.

êêêè A)ï22.62°êê B)ï20.87°

êêêè C)ï19.78°êê D)ïå of ç
@fig1511.bmp,25,118
ü
#êè (5.33)ìï=ï(12.6)ì + (8.9)ì - 2∙(12.6)∙(8.9)∙cos A

êêêë cos Aï≈ï.93437

êêêêïAï≈ï20.87°
Ç B
 8
êêêëFind the length of the tunnel from A to B.

êêêè A)ï447.407 mêëB)ï586.3 m

êêêè C)ï422 mêêïD)ïå of ç
@fig1512.bmp,25,118
ü
#êêïcìï=ï(253)ì + (373)ì - 2∙(253)∙(373)∙cos 89.1°

#êêêêïcìï≈ï200,173.0236

êêêêïcï≈ï447.407 m
Ç A
 9
êêêëFind angle B in the figure.

êêêè A)ï109.47°êêB)ï89.63°

êêêè C)ï93.8°êêïD)ïå of ç
@fig1513.bmp,25,118
ü
#êêë18ìï=ï10ì + 12ì - 2∙(10)∙(12)∙cos B

êêêëcos Bï≈ï-.3333

êêêêïBï≈ï109.47°
Ç A
 10êêFind the distance between points A and B which
êêêèare located on the opposite side of the river
êêêèfrom a forward observer.
êêêè A)ï79.6 mêê B)ï66.53 m

êêêè C)ï84.32 mêêD)ïå of ç
@fig1514.bmp,25,118
üëFirst, find CA using triangle ACD and the Law of Sines.
êêêëCAêè100 m
#êêêè───────è=è───────
êêêèsin 36°êsin 77°
êêêêCAè≈è60.325 m
êè Next, find CB using triangle CDB and the Law of Lines.
êêêëCBêè 100 m
#êêêè───────è=è───────
êêêèsin 80°êsin 58°
êêêêCBè≈è116.126 m
êè Now, find AB using triangle ABC and the Law of Cosines.
#ë (AB)ì = (60.325)ì + (116.126)ì - 2∙(60.325)∙(116.126)∙cos 25°
êêêë ABè≈è66.53 m

Ç B
äïPlease find the indicated part of the given figure if
êêpossible.
â

êêêèPlease see the Details.
éS
@fig1515.bmp,15,118
èIn the previous problems in this section, the three cases, an-
gle-side-angle, side-angle-side, and side-side-side, were covered.ïA
fourth case, when you know an angle and two sides with one side oppo-
site the given angle, will now be considered.ïThis fourth case has four
possible subcases.
êêêëIn the subcase, described by the figure, angle
êêêïA, side b, and side "a" are known.ïSince "a" is
êêêïless than the altitude, h = b∙sin A, there is no
êêêïtriangle solution to this problem.
êêêêêë0 solutions

êêêè In the second subcase angle A, side b, and
@fig1516.bmp,15,200
êêêïside "a" are known.ïSince side "a" is equal in
êêêïlength to the altitude, h = b∙sin A, there is
êêêïone right triangle solution.
êêêêêë1 solution

êêêëIn this case, as in the other cases, angle A,
@fig1517.bmp,15,200
êêêïside b, and side "a" are the known parts.ïHow-
êêêïever, since h < a < b, there are two distinct
êêêïtriangle solutions to this problem.
êêêêêë 2 solutions

êêêë In this case also, angle A, side b, and side
@fig1518.bmp,15,200
êêêè"a" are known.ïSince a > b, there is only one
êêêè triangle solution to this problem.
êêêêêë 1 solution


êêêè Since the given information is the same
êêêè in each one of the four cases, the first
êêêè step is to find the length of the altitude
êêêè and use this to determine which case is
êêêè involved.ïOnce you know which case is in-
êêêè volved, you can then use either the Law of
êêêè Sines or the Law of Cosines to find the
êêêè measure of unknown parts of the triangle.
 11
êè Find angle B, if possible, given that angle A is 35°, side b
is 12 inches, and side "a" is 2 inches.

êêèA)ï47° or 133°êê B)ï36°

êêèC)ï90°êêêïD)ïå of ç
üïIn this problem you are given one angle and two sides with one
side opposite the given angle.ïThe first step is to find "h", the
length of the altitude.
êêè hï=ïb∙sin Aï=ï12∙sin 35°ï≈ï6.88
Since side "a", which is 2 inches, is less than h, there is no triangle
solution to this problem.
Ç D
 12
êè Find angle B, if possible, given that angle A is 30°, side b
is 12 inches, and side "a" is 6 inches.

êêèA)ï55° or 125°êê B)ï58°

êêèC)ï90°êêêïD)ïå of ç
üïIn this problem you are given one angle and two sides with one
side opposite the given angle.ïThe first step is to find "h", the
length of the altitude.
êêè hï=ïb∙sin Aï=ï12∙sin 30°ï=ï6
Since side "a", which is 6 inches, is equal to h, there is one right
triangle solution to this problem.ïTherefore, angle B is 90°
Ç C
 13
êè Find angle B, if possible, given that angle A is 45°, side b
is 16√2 inches, and side "a" is 18 inches.

êêèA)ï62.74° or 117.26°êïB)ï83°

êêèC)ï90°êêêïD)ïå of ç
üïIn this problem you are given one angle and two sides with one
side opposite the given angle.ïThe first step is to find "h", the
length of the altitude.
êêè hï=ïb∙sin Aï=ï16√2∙sin 45°ï=ï16
Since side "a", which is 18 inches, is greater than h, and side "a" is
less than side b, there are two triangle solutions.
êêêè 18êè16√2
#êêêï───────è=è─────
êêêïsin 45°êsin B
êêêè sin Bè≈è.8889
Thus, B could be 62.74° or it's complement 117.26°.
Ç A
 14
êè Find angle B, if possible, given that angle A is 47°, side b
is 43.5, and side "a" is 52.1.

êêèA)ï56° or 124°êê B)ï37.63°

êêèC)ï90°êêêïD)ïå of ç
ü
êSince a > b in this problem, there is no need to find h.ïYou
know already that there is only one solution to this problem.ïYou
can use the Law of Sines to find angle B.
êêêë52.1ê 43.5
#êêêè───────è=è─────
êêêèsin 47°êsin B
êêêè sin Bè≈è.6106
êêêê Bè≈è37.63°
Ç B