Using the law of sines, how do you find the length of c to the nearest tenth of an inch?  This is an oblique triangle.  The angel opposite side c is 140degrees.  The only other defined parts are...

Using the law of sines, how do you find the length of c to the nearest tenth of an inch?

 

This is an oblique triangle.  The angel opposite side c is 140degrees.  The only other defined parts are the two sides adjacent to the 140degree angle, measuring 5inches and 9inches.


This appears to be a trick question with no solution as it cannot be solved with the law of sines--given that this is an SSA case and does not provide an angle opposite a measured side.  However, (to me anyway) it almost seems too easy of a trick question, so I would greatly appreciate a thorough review of the problem to make sure.  Thanks!

 

Asked on by kboy217

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lfryerda | High School Teacher | (Level 2) Educator

Posted on

Since the angle is between the two sides, this is not actually a SSA problem, but a SAS problem, which is one of the two cases that must be dealt with using the cosine law instead of sine law (the other case is SSS, where no angles are given).

The formula for cosine law is `c^2=a^2+b^2-2ab cos C` , where we are solving for c.

In this case, upon substitution, we get

`c^2=5^2+9^2-2(5)(9)cos 140`

`c^2=25+81-90(-0.7660)`

`c^2=174.944`   we can take square roots of both sides

`c=\sqrt(174.944)`

`c=13.2` 

Supposing that you can't use the cosine law in this case, it is still possible to get a solution to the problem.

Solution without cosine law

Draw the triangle so that the 5" side is horizontal and the 9" side is drawn up at a 140 degree angle.  Then from the top of the triangle, draw a new vertical line down to the horizontal line.  You now have a right-angled triangle that has another right-angled triangle inside it, with your original triangle also inside it

The small right-angled triangle has a 40 degree angle (since 140+40=180), and hypotenuse of 9".  By normal trig, this means that the hoizontal leg is given by `9cos 40 =6.8944`. The base of the large right-angled triangle is now `5+6.8944=11.8944`.  The vertical leg of the small triangle is `9 sin40=5.7851`, which is the same as the vertical leg of the large right-angled triangle.

By the Pythagorean theorem, we can find the hypotenuse of the large triangle, since we know the legs of the triangle.

This means that `c^2=11.8944^2+5.7851^2` , which means that `c=\sqrt{171.944}=13.2`.

The second method still does not require the sine law, which is really inapplicable to this question.

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