Asked by Tyler
Use the Law of Sines to solve each triangle.
P=110 degrees p=125, r=200 in triangel PQR
P=110 degrees p=125, r=200 in triangel PQR
Answers
Answered by
Steve
p/sinP = r/sinR
125/sin 110 = 200/sinR
133.02 = 200/sinR
sinR = 1.50
eh? sine is always less than 1.
And, since P = 110 degrees, the other angles must be less than 70 degrees.
Since p is opposite the largest angle, it must be the longest side. How then can r be 200?
Better check your problem for typos.
125/sin 110 = 200/sinR
133.02 = 200/sinR
sinR = 1.50
eh? sine is always less than 1.
And, since P = 110 degrees, the other angles must be less than 70 degrees.
Since p is opposite the largest angle, it must be the longest side. How then can r be 200?
Better check your problem for typos.
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