Solve Triangle ABC: A=50, b=15, c=11

[asura]

Homework Statement

Solve triangle ABC...
A = 50 degrees
b= 15
c = 11

Homework Equations

Law of Cosine and Law of Sine

The Attempt at a Solution

since I have two sides and the angle in between, I used law of cosine to find a
a² = 30² + 15² - 2(30)(15)cos 50
a = 23.38

then since I have angle A and the opposite side a, I used law of sine to find the other two angles

sin C / 30 = sin 50 / 23.38 = sin B / 15

the problem with this is that I get two different values of B and C

if I solve for C, then C = 79.4 and B = 50.6 (180 - 50 - 79.4)
if I solve for B, then B = 29.4 and C = 100.6

which answer would be correct?

 

[kidmode01]

Unless I'm really confused we have A = angle between b and c and a is the length opposite of the angle A. Then law of cosines says:

a^2 = b^2 + c^2 - 2*(a)*(b)*cos(A)
a^2 = 15^2 + 11^2 - 2*(11)(15)cos(50)
a = 11.57065637

 

[Mark44]

Homework Statement

Solve triangle ABC...
A = 50 degrees
b= 15
c = 11

Homework Equations

Law of Cosine and Law of Sine

The Attempt at a Solution

since I have two sides and the angle in between, I used law of cosine to find a
a² = 30² + 15² - 2(30)(15)cos 50

You were given that b = 15 and c = 11, but in your formula you used c = 30.

a = 23.38

then since I have angle A and the opposite side a, I used law of sine to find the other two angles

sin C / 30 = sin 50 / 23.38 = sin B / 15

the problem with this is that I get two different values of B and C

if I solve for C, then C = 79.4 and B = 50.6 (180 - 50 - 79.4)
if I solve for B, then B = 29.4 and C = 100.6

which answer would be correct?

 

[asura]

wow i copied the question wrong, c is supposed to be 30
for some reason i put 11...

but my question is still the same

which answer is correct

 

[Mark44]

Homework Statement

Solve triangle ABC...
A = 50 degrees
b= 15
c = 30

Homework Equations

Law of Cosine and Law of Sine

The Attempt at a Solution

since I have two sides and the angle in between, I used law of cosine to find a
a² = 30² + 15² - 2(30)(15)cos 50
a = 23.38

then since I have angle A and the opposite side a, I used law of sine to find the other two angles

sin C / 30 = sin 50 / 23.38 = sin B / 15

the problem with this is that I get two different values of B and C

if I solve for C, then C = 79.4 and B = 50.6 (180 - 50 - 79.4)
if I solve for B, then B = 29.4 and C = 100.6

which answer would be correct?

(I changed your value for c, above.)
My guess is that C = 100.6 degrees is correct. From your work in finding a, you should be able to verify this by using the Law of Cosines again. This time you know b, a, and c, and you can use it to find B.

 

[asura]

i appreciate the response, but that sounds like guess and check
is there a method that will always show you the correct answer?

 

[Tedjn]

The longest side of the triangle is always opposite the largest angle.

 

[naresh]

i appreciate the response, but that sounds like guess and check
is there a method that will always show you the correct answer?

Yes, using the cosine law will always give you a "correct answer" because the angle lies between 0 and 180 degrees. The sine law is fine, but remember that sin(79.4) = sin(100.6), and you need to be careful with the inverse sine. [In general sin(90-x) = sin(90+x)]. You will need to use something else to verify which one is correct (like largest angle opposite longest side).

 

[HallsofIvy]

What do you have against "guess and check"?