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February 25th, 2017, 01:47 PM   #1
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Find Area of Triangle

Hi,

Could anyone provide a complete solution to the attached question?

I calculated 400 but the teacher got 463.



Thanks in advance for any help.

Best regards,
Joseph

Last edited by skipjack; August 7th, 2017 at 10:42 PM.
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August 7th, 2017, 10:57 PM   #2
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Calculate angle C using the sine rule, then calculate angle A.

Area = (1/2)AB × AC × sin(A) = 463 cm² approximately.
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September 1st, 2017, 03:40 PM   #3
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I used the law of sines to calculate one unknown angle and got the other unknown angle by doing 180 - the sum of the other two angles. Then I used the law of sines to calculate the unknown side. Then I used the semiperimeter formula for area given the three sides. I wasn't working with whole numbers, but I got a number close to 463.
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September 24th, 2017, 05:03 AM   #4
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This problem is not "well posed"- there exist two distinct triangles having the given data. While there is an "SAS" congruence rule, there is no "SSA" (the given angle not between the two given sides as here).

Consider this: Draw a line and mark off an interval of length 25 cm. At one end of the interval construct an angle of 75 degrees and extend that side indefinitely. At the other end of the 25 cm interval, use compasses to mark an arc of radius 40 cm. There are three possibilities. (1) The radius might not be long enough- the arc does not cross the other side at all- there is no such triangle. (2) The radius is exactly long enough- the arc is tangent to the other side- there is a single (right) triangle. (3) The radius is too long- the arc crosses the other side in two different places- there are two distinct triangles.

We can, as skipjack suggests, use the sine law to determine angle C, but it is not determined uniquely. $\displaystyle \frac{\sin(75)}{40}= \frac{\sin(C)}{25}$ so $\displaystyle \sin(C)= \frac{25 \sin(75)}{40}= 0.6037$. Now $\displaystyle C= \sin^{-1}(.6037)= 37.1$ degrees but the sine function is not "one to one" so does not have a unique inverse. $\displaystyle C= 180- 37.1= 142.9$ degrees gives us a second triangle. I suppose you can argue that the triangle in the picture is clearly an acute triangle but that is why you can get two different areas for the triangle.

Last edited by skipjack; September 27th, 2017 at 03:00 PM.
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September 24th, 2017, 06:15 PM   #5
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I understand that SSA doesn't provide only one possibility for the shape. However, you don't have to guess that the triangle is acute because you know it is acute. If you drew a vertical line perpendicular to AC that goes through A, it will produce a right angle that is larger than angle A, and the same is true for drawing a perpendicular line going through C producing a right angle greater than angle C. In addition, this problem gives a 75 degree angle, and a triangle cannot have a 75 degree angle and a 142.9 degree angle.
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September 28th, 2017, 03:36 PM   #6
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Quote:
Originally Posted by Country Boy View Post
. . . and extend that side indefinitely.
It's appropriate to extend it in just one direction, so when the radius is too large (your third case), the arc may intersect the extended side in just one place. As Evanj correctly points out, the "ambiguous case" of SSA doesn't arise for the given data. Many textbooks fail to explain how best to determine whether ambiguity arises. For example, it can't arise if the given angle is obtuse or a right angle, but many books don't draw attention to this.
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