Trigonometry/ProOneSoftware-Trigonometry-Win31.iso/trig/chapter1.3r

Text File | 1995-04-09 | 7KB | 220 lines

 218
à 1.3ïSolving Right Triangles

äïPlease find the indicated parts of the following right
êëtriangles.
âêêêFind side "x" in this triangle.

êêêêêïsin 33.5°è=èx/20ft

êêêêë(20ft)∙(.552)è≈èx
êêêêêè11.04 ftè≈èx
@fig1301.bmp,15,100

éSïTo solve a right triangle means to find the lengths of the
sides and the measures of the angles.ïAll parts of a right triangle
can be found if you know the length of two sides or if you know the
length of one side and the measure of one of the acute angles.ïBasi-
cally, the tools that you have to work with to solve an arbitrary right
triangle are the Pythagorean Theorem, the trigonometric ratios, the in-
verse trigonometric functions, and the fact that the two acute angles
are complementary.
@fig1301.bmp,15,160
êêêIn the example, we know one side and one of the
êêêacute angles.ïYou should choose a trigonometric
êêêfunction that involves the given angle, the given
êêêside, and the unknown side "x".ïThere are two
êêêtrigonometric functions that we could use in this
êêêproblem.ïThey are the "sin" and the "csc".ïIf you
êêêchoose the "sin" function, then you have the
êêêfollowing equation.

êêêèsin 33.5°è=èx/(20 ft)
êêêè (20 ft)∙(.552)è≈èx
êêêê11.04 ftè≈èx

Thus, the unknown side is 11.04 ft.ïIf you wanted to find other parts
of this triangle, you could subtract 33.5° from 90° to get the other an-
gle, and you could use the Pythagorean Theorem to get the third side.
There will usually be more than one correct way to find the parts.ïYou
will notice that many times the answers are approximate.ïThis is be-
cause the ratios and angles that we get from the calculator are neces-
sarily rounded off.ïYou should round your answers only once if possi-
ble at the end of the problem.ïThe following definitions will be help-
ful when solving applications.
1)ïAngle of Elevation - An angle of elevation is the angle formed by
looking at the horizon line then looking up to a point above the
horizon.
2)ïAngle of Depression - This is the angle formed by looking at the
horizon line then looking down to a point below the horizon.
3)ïBearing - A bearing is a path or direction described by using North
or South as a base line then indicating an acute angle either East
or West of the base line.ïFor example, S 30°W means a path in the di-
rection of 30° to the West of South.
4)ïHeading - A heading is an angle between 0° and 360° measured clock-
wise from North.
 1êëFind side "x" in this figure.

êêêè A)ï36.7 ydêêB)ï24.58 yd

êêêè C)ï28.53 ydêë D)ïå of ç

@fig1302.bmp,15,118
ü
êêêcos 52.3°è=è x/(40.2 yd)

êêê (40.2 yd)∙(.6115)è≈èx

êêêë 24.58 ydè≈èx
Ç B
 2êëFind angle A in this figure.

êêêè A)ï28.7°êêïB)ï31.89°

êêêè C)ï35.62°êê D)ïå of ç

@fig1303.bmp,15,118
ü
êêêïtan Aè=è(18.7)/(26.1)

êêêë tan Aè≈è.7165

#êêêëAè≈ètanúî .7165
êêêê Aè≈è35.62°
Ç C
 3êëFind side "x" in this figure.

êêêè A)ï24.01 cmêë B)ï27.5 cm

êêêè C)ï26.23 cmêë D)ïå of ç

@fig1304.bmp,15,118
ü
#êêêï(28.1)ìè=èxì + (14.6)ì

#êêêï(28.1)ì - (14.6)ìè=èxì

#êêêê576.45è=èxì
êêêë 24.01 cmè≈èx
Ç A
 4êëFind the height of the tree.

êêêè A)ï196.26 ftêëB)ï58.7 ft

êêêè C)ï64.2 ftêêD)ïå of ç

@fig1305.bmp,15,118
üïThe height of the tree can't be measured directly, however, you
can measure along the ground from the base of the tree a distance of
100 ft.ïAlso, from ground level you can measure an angle of elevation
of 63° to the top of the tree.
êêêètan 63°è=èh/(100 ft)
êêêï(100 ft)∙(1.9626)è≈èh
êêêë196.26 ftè=èx
Ç A
 5êëFind the height of the cinder cone.ïAngle A is
êêê47.3°, angle B is 56.9°, and AB is 500 ft.

êêêè A)ï868.5 ftêë B)ï763.4 ft

êêêè C)ï1,877.61 ftêèD)ïå of ç
@fig1306.bmp,15,118
üïIn this problem both the height and the distance from A to C are
inaccessible.ïYou can, however, measure the angle of elevation at B to
be 56.9°, the distance from A to B to be 500 ft, and the angle of eleva-
tion at A to be 47.3°.ïYou then have two right triangles involving the
unknown height "h".
êè tan 47.3°è=èh/(x + 500), and tan 56.9°è=èh/x
êï(tan 47.3°)∙(x + 500)è=èh, and (tan 56.9°)∙xè=èh
êê (tan 47.3°)∙(x + 500)è=è(tan 56.9°)∙x
êë500∙(tan 47.3°)è=èx∙(tan 56.9° - tan 47.5°)
êêê 541.845è≈è(.4427)∙x
êêêë1,224 ftè≈èx
This value of "x" can be substituted back in to find "h".
êê hè=èx∙tan 56.9°è≈è1,877.61 ft
Ç C
 6êëFind the height of the lighthouse.

êêêè A)ï106.7 ftêë B)ï145.2 ft

êêêè C)ï152.3 ftêë D)ïå of ç

@fig1307.bmp,15,118
üïThis time the height of the lighthouse and the distance to the
lighthouse are inaccessible.ïYou can, however, measure the angle of
elevation at A to be 43.1°, the distance from A to B to be 100 ft, and
the angle at B to be 57.2°.ïNext, you should use the flat triangle to
find side "x".
êêê tan 57.2°è=èx/(100 ft)
êêêè 155.17 ftè≈èx
Then, use the verticle triangle to find side "h".
êêê tan 43.1°è=èh/(155.17 ft)
êêêë 145.2 ftè≈èh
Ç B
 7êëFind the depth of the canyon.ïThe angle of depres-
êêêsion at A is 62.3°, AB is 100 ft, and angle B is
êêêè71.5°.
êêêè A)ï264.62 ftêëB)ï369.2 ft

êêêè C)ï186.1 ftêë D)ïå of ç
@fig1308.bmp,15,118
üïIn this problem the canyon depth is inaccessible.ïThe angle of
depression from A down to the bottom of the canyon is 62.3°, AB is
100 ft, and the angle at B is 71.5°.ïYou should first find side "x".

êêê tan 71.5°è=èx/(100 ft)
êêêè 298.87 ftè≈èx
Then, use the other triangle to find side "h".
êêê sin 62.3°è=èh/(298.87 ft)
êêêë264.62 ftè≈èh
Ç A
 8ïTwo towers are 12.85 miles apart.ïA target is spotted by
the first tower at a bearing of N 23°42' W.ïIt is also sighted
by the second tower at a bearing of N 66°18' E.ïIf the first tower
is due East of the second tower, find the distance from the first
tower to the target.
êêèA)ï6.13 miêê B)ï5.17 mi
êêèC)ï7.62 miêê D)ïå of ç
üêê Since angle A and angle B add up to 90°, angle C
êêêmust be 90°.ïThe first tower is located in the
êêêfigure at B, and the angle at B is 66°18'.ïThis
êêêgives us one side and one acute angle in the right
êêêtriangle ABC.
êêêêë cos 66.3°è=èCB/(12.85 mi)
@fig1309.bmp,50,980
êêêêë 5.17 milesè=èCB
Ç B
 9ïA small airplane travels at a heading of 212° for 120
minutes at an average speed of 95 miles per hour.ïIt then turns
to a heading of 122° and flies for 175 minutes at the same speed.
If there is no appreciable wind to affect the speed of the plane,
find the distance from the plane to its original starting point.
êë A)ï335.10 miêè B)ï201.4 mi
êë C)ï186.4 miêëD)ïå of ç
üêê Notice that angle C must be a right angle.
êêêSince triangle ABC is a right triangle, you can
êêêfind the length of the legs and use the Pytha-
êêêgorean Theorem to find the distance AB.

êêêAC = rate∙time = 95 mi/hr∙2 hr = 190 mi
@fig1310.bmp,50,980
êêêBC = rate∙time = 95 mi/hr∙2.917 hr = 277.115 mi
#êêê (AB)ìï=ï(190)ì + (277.115)ì
êêêèABï=ï335.10 miles
Ç A
 10êè Find the distance, "x", across the lake.

êêêèA)ï983.4 ftêè B)ï1,286.4 ft

êêêèC)ï1,760.47 ftê D)ïå of ç

@fig1311.bmp,15,118
ü

êêêè cos 73.5°è=è(500 ft)/x

êêêèxè=è(500 ft)/(cos 73.5°)

êêêêxè=è1,760.47 ft
Ç C