Oblique Triangles
& Law of Sines
& Law of Sines
Find the length of the side a. Do not use a calculator.
<A=60°, <B= 75°, c=√2
C=45°
√2sin60/sin45=1.7
1.7=√3
Answer: √3
Find the length of c, to the nearest 10th of an inch.
∆ABC, a=80in, <C=136°, <A=16°
<B=180-136-16=28°
c/(sin136 )=80/sin16
c=80sin136/sin16=201.615
Answer: 201.6
Find the length of a, to the nearest 10th of an inch.
∆ABC, b=1.3in, c=3.5in, A=159°
a^2=1.3^2+3.5^2-2(1.3)(3.5)cos159
a^2=1.69+12.25-2(1.3)(3.5)(-0.93358)
a^2=1.69+12.25+8.49558
a^2=22.43558
a=√22.43558=4.737
Answer: 4.7
Find the dot product for the pair of vectors.
<2,4>,<0,-1>
(2)(0)+(4)(-1)=0
0-4
Answer: -4
Solve the problem: To build the pyramids they used causeways to transport materials to the site. One of them was 3000 feet long, with a slope of about 2.3°. How much force would be required to pull a 60-ton monolith along this causeway.
sin2.3°=x/60
60sin2.3°=2.4
Answer: 2.4 tons
Determine the remaining sides and angles of ∆ABC.
<C=115.5°, <A= 27.2°, c=76ft
<B=180-142.7=37.8°
a=76sin27.2°/sin115.5°=38.48ft
b=76sin37.3°/sin115.5°=51.02ft
Answer: <B=37.8°, a=38.48ft, b=51.02ft
Find the length of b, to the nearest 10th of an inch.
∆ABC, b=9.8in, <C=106°, <A=17°
<B=180-106-17=57°
c/(sin106 )=9.8/sin57
c=9.8sin106/sin57=11.232
Answer: 11.2
Find the length of c, to the nearest 10th of a centimeter.
∆ABC, a=8.2cm, b=7.1cm, <C=7°
c^2=8.2^2+7.1^2-2(8.2)(7.1)cos7
c^2=67.24+50.41-2(8.2)(7.1)(0.992546)
c^2=67.24+50.41-115.57207
c^2=2.07793
c=√2.07793=1.442
Answer: 1.4
Find the magnitudes of the horizontal and vertical components of v, if α is the angle of inclination of v from the horizontal.
α=27°30',|v|=15.4
|x|=15.4cos27°30'=13.7
|y|=15.4sin27°30'=7.11
Answer: 13.7, 7.11
Solve the problem: Two forces of 692 newtons and 432 newtons act at a point. The resultant force is 786 newtons. Find the angle between the forces.
786^2=692^2+423^2-2(692)(423)cos(180-θ)
786^2-692^2-423^2/-2(692)(423)=.0683204881
cos^-1(.068320)=86.08
180-86.08=93.9°
Answer: 93.9°
Find the area of the triangle using the formula A=1/2bh and verify that the formula A=1/2absinC gives the same result.
√A=60°, √C= 90°, c=2, b=1, a=√3
1/2 (1)(√3)
1/2(√3)=√3/2
√3/2=.866
1/2(√3)(1)sin90°=.866
Answer: √3/2
Find all possible values of <A, to the nearest degree.
∆ABC, a=92in, c=28in, <C=8°
<A/92=sin8/28
sinA=92sin8/28=.457283
Q2: 180-27=153°
8+27=35°
8+153=161°
Answer: 27° and 153°
Find the measure of <B to the nearest degree.
∆ABC, a=9.6cm, b=7.4cm, c=9.9cm
cosB=9.9^2+9.6^2-7.4^2/2(9.9)(9.6)
cosB=135.41/190.08=.7123843
B=cos^-1(.7123843)=44.571=45°
Answer: 45°
Find the angle between each pair of vectors.
≺3,4≻ ≺0,1≻
uv=(3)(0)+(4)(1)=4
u=√3^2+4^2=5
v= √0^2+1^2=1
(5)(1)=5
cos^-1(4/5)=36.87°
Answer: 36.87°
Solve the problem: An airline route from Chicago to New Jersey is on a bearing of 233°. A jet flying at 450mph on that bearing flies into a wind blowing at 39mph from a direction of 114°. Find the resulting bearing and groundspeed of the plane.
233°=450mph
114°=39mph
233-114=119°
39^2+450^2-2(39)(450)cos119°
√221037.8177=470
39sin119°/470=4°
4°+223°=237°
Answer: 237°; 470mph
Find the area of ABC given
A=42.5°, b=13.6m, a=10.1m
A=1/2(13.6)(10.1)(sin42.5)
A= 46.4m^2
Find all possible values of <C, to the nearest degree.
∆ABC, c=45cm, b=21cm, <B=64°
sinC/45=sin64/21
sinC=45sin64/21=1.9259872
C=sin^-1(1.9259872)= ERROR
Answer: No Possible Triangles
Find the measure of <C to the nearest 10th of a degree.
∆ABC, a=880in, b=140in, c=770in
cosC=880^2+140^2-770^2/2(880)(140)
cosC=201100/246400=.8161526
C=cos^-1(.8161526)=35.2985=35.3°
Answer: 35.3°
Two forces act at appoint in the plane. The angle between the two forces is given. Find the magnitude of the resultant force.
Forces of 116 and 139 pounds, forming an angle of 140°50'
180-140°50'=39°10'
|→v^2|=116^2+139^2-2(116)(139)cos39°10'
|→v^2|=√7774.736777=88.2
Answer: 88.2lbs
Solve the problem: A plane is headed due south with an airspeed of 192mph. A wind from a direction of 78° is blowing at 23mph. Find the groundspeed and resulting bearing of the plane.
180°-78°=102°
|v|^2=23^2+192^2(23)(192)cos102°
v=√39,229.28
v=198
23sin102°/198=.1135
sin^-1(.1135)=6.5°
180°+6.5°=186.5
Answer: 198mph; 186.5°
Find the area of ABC given
C=72.2°, b=43.8ft, a=35.1ft
A=1/2(35.1)(43.8)(sin72.2)
A= 731.9ft^2
Find the area of ∆ABC, to the nearest 10th of a square inch.
∆ABC, b=240in, <A=45°, <B=75°
<C=180-45-75=60°
a/(sin60)=240/sin75
a=240sin60/sin75=215.1781133
Area=1/2(240)(215.1781133)sin45=18258.468
Answer: 18258.5
Find the area of ∆ABC,
a=25.4yd, b=38.2yd, c=19.8yd
s=1/2(25.4+38.2+19.8)=41.7
A=√41.7(41.7-25.4)(41.7-38.2)(41.7-19.8)
√41.7(16.3)(3.5)(21.9)
Answer: 228yd^2
Find the magnitude and direction angle (to the nearest tenth) for each vector. Give the measure of the direction angle as an angle in [0,360°).
≺8√2, -8√2≻
Quadrant 4 (+,-)
M=√(8√ 2)^2+(-8√ 2)^2
√128+128=16
tan^-1(-8√2/8√2)=-45
360-45=315
Answer: 315°
Solve the problem: A ship sailing due east in the North Atlantic has been warned to change course to avoid a group of icebergs. The captain turns and sails on a bearing of 62° for a while, then changes course again to a bearing of 115° until the ship reaches its original course. How much farther did the ship have to travel to avoid the icebergs?
90°-62°=28° 180-????
360°-115°-118°=127°
180°-127°-28°=25°
50sin28/sin127=29.4
50/sin127 x sin25=26.5
26.5+29.4=55.9
55.9-50=5.9
Answer: 5.9 miles