Expert: Sherman D. - 4/29/2009

Question
hiii my name is ajanthana and i`m rather confused. I used the quadratic formula but i keep getting the wrong answer whcih si why i need your help, id really appreicate it thank you :) !

using the cosine law, find the length of side c. How many triangles are possible? Explain.
angle a= 60º
length of a is 5√3 cm (5 ROOT 3)
length of b is 10 cm
angle b is unknown
length of c and angle of c is also unknown

suppose that the length of side a is changed to each of the following values. By using the cosine law to find the length of c, determine how  many triangles are possible. Explain your reasoning.
a) 9 cm     b) 12 cm   c) 8 cm


Answer
(5sqrt(3))^2 = c^2 + 10^2 - 2(10 * c)cos(60)
75 = c^2 + 100 - 20c*(1/2)
-25 = c^2 - 10c
c^2 - 10c + 25 = 0
(c - 5)^2 = 0
c = 5

100 = 25 + 75 - 2(25sqrt(3))cos(B)
100 = 100 - 50sqrt(3)cos(B)
0 = -50sqrt(3)cos(B)
cos(B) = 0
B = 90

now since 180 - 90 - 60 = 30, lets see if we can get it to work out.

5^2 = 10^2 + (5sqrt(3))^2 - 2(10 * 5)cos(30)
25 = 100 + 75 - 100(sqrt(3)/2)
25 = 175 - 100sqrt(3)/2
-150 = -100sqrt(3)/2

as you can see both sides are equal, and therefore there are no triangles possible to be made from the info you have given me.

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a.)
9^2 = c^2 + 10^2 - 2(c * 10)cos(60)
81 = c^2 + 100 - 20c*(1/2)
-19 = c^2 - 10c
c^2 - 10c + 19 = 0

(10 +/- sqrt(100 - 4(19 * 1)))/(2(1))
(10 +/- sqrt(100 - 76))/2
(10 +/- sqrt(24))/2
(10 +/- 2sqrt(6))/2
c = 5 +/- sqrt(6)

100 = (5 +/- sqrt(6))^2 + 81 - 2(9(5 +/- sqrt(6))cos(B)
19 - (5 +/- sqrt(6))^2 = -18(5 +/- sqrt(6))cos(B)
cos(B) = (19 - (5 +/- sqrt(6))^2)/(-18(5 +/- sqrt(6)))
B = about 74.207 or 105.793

C = about 45.793 or 14.207

2 possible triangles

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b.)
12^2 = c^2 + 10^2 - 2(10c)(1/2)
144 = c^2 + 100 - 10c
c^2 - 10c - 56 = 0

c = (10 +/- sqrt(100 - 4(1 * -56)))/(2(1))
c = (10 +/- sqrt(100 + 224))/2
c = (10 +/- sqrt(324))/2
c = (10 +/- 18)/2
c = (-8/2) or (28/2)
c = -4 or 14

since you can't have a negative length

c = 14

14^2 = 144 + 100 - 2(12 * 10)cos(C)
196 = 244 - 240cos(C)
-48 = -240cos(C)
cos(C) = 48/240
C = about 78.463

B = 41.537

One Triangle is possible

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c.)
8^2 = c^2 + 100 - 10c
64 = c^2 - 10c + 100
c^2 - 10c + 36 = 0

c = (10 +/- sqrt(100 - 4(1 * 36)))/(2(1))
c = (10 +/- sqrt(100 - 144))/2
c = (10 +/- sqrt(-44))/2
c = (10 +/- 2isqrt(11))/2
c = 5 +/- isqrt(11)

since you can't have an imaginary length

unless i did something wrong somewhere, no triangles are possible.