Law of sine and cosine problem

• May 8th 2010, 02:28 PM
jwu
Law of sine and cosine problem
Just a simple problem. But I got stuck on some points.
We just learned the law of sine and the law of cosine from class.
And here's the assignment according this two topics:

Solve each triangle. Give lengths to three significant digits and angle measures to the nearest tenth of a degree.
a=8, b=5, ∠C=60°

My solution:
Because we have the condition "SAS". So I used the law of cosine to solve for c.
c²=a²+b²-2abcos 60°
=7

Then all we need to do next is to find the two angles of the triangle, right?
So I used the sine law：
sin C/c=sin A/a
sin 60°/7=sin A/8
sin A= sin 60°*8/7
A= 81.8° and 98.2°

Here's where I got stuck. The answer at the very end of book only shows the answer 81.8°. But I think 98.2° is also acceptable. There should be two possiblities. Can somebody tell me why it doesn't fit?
• May 8th 2010, 04:02 PM
masters
Quote:

Originally Posted by jwu
Just a simple problem. But I got stuck on some points.
We just learned the law of sine and the law of cosine from class.
And here's the assignment according this two topics:

Solve each triangle. Give lengths to three significant digits and angle measures to the nearest tenth of a degree.
a=8, b=5, ∠C=60°

My solution:
Because we have the condition "SAS". So I used the law of cosine to solve for c.
c²=a²+b²-2abcos 60°
=7

Then all we need to do next is to find the two angles of the triangle, right?
So I used the sine law：
sin C/c=sin A/a
sin 60°/7=sin A/8
sin A= sin 60°*8/7
A= 81.8° and 98.2°

Here's where I got stuck. The answer at the very end of book only shows the answer 81.8°. But I think 98.2° is also acceptable. There should be two possiblities. Can somebody tell me why it doesn't fit?

Hi jwu,

You were initially given two sides and an included angle.
You used the law of cosine to find side C.

So, now you have 3 fixed lengths and one angle.

The ambiguous case comes into play when you are given
two sides and an angle opposite one of them.

This is obviously not the case here.
• May 8th 2010, 04:10 PM
Quote:

Originally Posted by jwu
Just a simple problem. But I got stuck on some points.
We just learned the law of sine and the law of cosine from class.
And here's the assignment according this two topics:

Solve each triangle. Give lengths to three significant digits and angle measures to the nearest tenth of a degree.
a=8, b=5, ∠C=60°

My solution:
Because we have the condition "SAS". So I used the law of cosine to solve for c.
c²=a²+b²-2abcos 60°
=7

Then all we need to do next is to find the two angles of the triangle, right?
So I used the sine law：
sin C/c=sin A/a
sin 60°/7=sin A/8
sin A= sin 60°*8/7
A= 81.8° and 98.2°

Here's where I got stuck. The answer at the very end of book only shows the answer 81.8°. But I think 98.2° is also acceptable. There should be two possiblities. Can somebody tell me why it doesn't fit?

Hi jwu,

another way to look at this is...

if you draw the diagram, evenly roughly will do,
notice that the triangle's shape is fully determined
after you've drawn the 2 sides and the included angle.

All that's left is to join the remaining two endpoints.
There's no other way to draw this triangle.

Hence the remaining angles are already fixed.

You can verify side "a" with the law of cosines.

$a^2=b^2+c^2-2bccosA$

$8^2=5^2+7^2-2(5)7cosA$

$64=25+49-70cosA$

$70cosA=74-64=10$

$cosA=\frac{1}{7}$

$A=cos^{-1}\left(\frac{1}{7}\right)=81.78^o$

or $360^o-81.78^o!!$ which we can certainly discard!