Sine Law
(Knowledge/Diagram)
(Knowledge/Diagram)
Sine Law
(Word Problems)
(Word Problems)
Cosine Law
(Knowledge/Diagram)
(Knowledge/Diagram)
Cosine Law
(Word Problems)
(Word Problems)
100
Is it true that in the formula a/sinA = b/sinB = c/sinC (sine law), B is the angle opposite of side a?
No, B is the angle opposite of side b while A is the angle opposite of side a.
100
Diana and James are standing at the seashore 10 miles apart. The coastline is a straight line between them. Both can see the ship in the water. The angle between the coastline and the line between the ship and Diana is 35 degrees. The angle between the coastline and the line between the ship and James is 45 degrees. How far is the ship from James?
Find 3 angles with angle theorem: 35+ 45
180 - 80 = 100º
10/sin100º = x/sin35º
x=10sin35º/sin100º
x=5.8 m
Therefore the ship is 5.8 m from James
100
What are the conditions we should have to solve a triangle with cosine law? (require 2 answers for points) (20s)
1. Two sides and a contained angle
2. All three sides
100
Tom, Ricky, and Harry are camping in their tents. If the distance between Tom and Ricky is 153 ft, the distance between Tom and Harry is 201 ft, and the distance between Ricky and Harry is 175 ft, what is the angle between Ricky, Harry, and Tom?
1532=2012+1752-2(201)(175)cosH
23409=40401+30625-70350cosH
-47617=-70350*cosH
H=47.4º
Therefore the angle is 47.4º
200
In the pictured triangle, ∠A is 98 degrees and ∠B is 12 degrees. If side a is 84 units long, approximately how long is side b?
18 units
200
3 members of one direction, Harry, Niall, and Louis, decide to try out a new formation on stage. They decide to use Harry's favourite shape, a triangle. The distance from Louis to Harry is 8ft, and the distance from Harry to Niall is 7 ft. the angle formed at Niall is 70º. What is the angle formed at Louis?
sinL/7=sin70º/8
L = sin-1(7(sin70º)/8)
L = 55.3º
Therefore Louis forms a 55.3º angle.
200
Solve for B to the nearest tenth of a degree
192=82+142-2(8)(14)cosX
361= 64+196-224cosX
-101=-224cosX
cosX=101/224
X=cos-1(101/224)
X=63.2º
200
Two ships leave port at 4 p.m. One is headed at a bearing of N 38º E and is traveling at 11.5 miles per hour. The other is traveling 13 miles per hour at a bearing of S 47º E. How far apart are they when dinner is served at 6 p.m.?
180-38-47=95
a=11.5(2)=23 miles, b=13(2)=26 miles
d2=232+262-2(23)(26)cos95º
d2=1205-1196cos95º
d=36.2
Therefore the ships are 36.2 miles apart at 6 pm.
300
Find the obtuse angle (to the nearest tenth of a degree) such that:
sinX/12 = sin15º/7
X= sin-1(12(sin15º)/7)
X = 26.3º
Obtuse angle: 180º-X
180º-26.3º=153.7º
X = 153.7º
300
The trumpet, clarinet, and flute sections formed a triangle during marching band practice. The clarinet line made a 34° angle with the flute line. The flute line made a 107° angle with the trumpet line. If the flute line was 27 feet long, find the length of the clarinet line.
x/sin107º=27/sin39º
x=41.0 ft
Therefore the length of the clarinet line is 41 ft.
300
Ｗhat is the unknown angle shown in the diagram?
24202=50502+60002-2*5050*6000*cosθ
5856400=61502500-60600000*cosθ
-55646100=-60600000*cosθ
<θ=23.34º
300
To approximate the length of a lake, a surveyor starts at one end of the lake and walks 245 yards. He then turns 110º and walks 270 yards until he arrives at the other end of the lake. Approximately how long is the lake? (Hint: He walked the two sides of a triangle.)
x2=2702+2452-2(270)(245)cos70º
x2=132925-132300cos70º
x=296.1
Therefore the lake is 296.1 yards long.
400
Triangle PQR has ∠P = 63.5° and ∠Q = 51.2° and r = 6.3 cm. What are the side lengths of p and q?
p = 6.21cm
q = 5.40 cm
400
A cottage under construction is to be 12.6 m wide. The two sides of the roof are to be supported by rafters that meet at an angle of 50 degrees. How long should the rafters be if they are the same length?
∠A + ∠C + 50 degrees = 180 degrees
2x + 50º = 180º
2x = 180 - 50
2x = 130
x = 65º
a/sinA = b/sinB
a/sin65º = 12.6/sin50º
A = 12.6(sin65)/ sin50
A =14.9 m
Therefore the rafters should be 14.9 m in length.
400
62=3.52+42-2(3.5)(4)cosθ
36=12.25+16-28cosθ
7.75=-28cosθ
<θ=106º
106-90=16º (how far from the vertical)
A:16º
400
Two ships leave the harbour at the same time. One ship travels on a bearing of S12°W at 14 miles per hour. The other ship travels on a bearing of N75°E at 10 miles per hour. How far apart will the ships be after three hours? Round to the nearest tenth of a mile.
x2=302+422-2(30)(42)cos117º
x=61.7
Therefore the ships are 61.7 miles apart after 3 hours.
500
In the diagram below, angle WZY is 60 degrees, angle XWY is 120 degrees, WX = 7m and WY = 8m. Find the length YZ.
180-120=60º
180-60-60=60º (WZY is an equilateral, so all the sides are 8m)
Ans:8m
500
A wall that is 1.4 m long has started to lean and now makes an angle of 80o with the ground. A 2 m board is jammed between the top of the wall and the ground to prop up the wall. What angle does the board make with the ground?
2/sin80º = 1.4/sinS (any variable is fine)
S = 1.4sin80º / 2
S = sin-1 (1.4sin80º / 2)
S = 44º
Therefore the board makes a 44º angle with the ground.
500
c2= 272+602-2(27)(60)cos121º
c2=4329-3240(-0.5150)
c2=5997.7234
c=77.4450mm
500
A plane leaves airport A and travels 580 miles to airport B on a bearing of N34°E. The plane later leaves airport B and travels to airport C 400 miles away on a bearing of S74°E. Find the distance from airport A to airport C to the nearest tenth of a mile.
x2=5802+4002-2(580)(400)cos108º
x2=496400-464000cos108º
x=799.9
Therefore the distance from airport A to airport C is 799.9 miles