Asked by Amone
In obtuse triangle PQR, P=51 degrees, p= 10cm, and the longest side, q=12cm.Draw the triangle and solve for Q to the nearest degree.
I did,
10/sin 51=12/sin Q
10(sin Q)/10=12(sin 51)/10
Q= 2nd function sin 0.9325751
Q=68 degrees
Q=180-68
Q=111 degrees
I did,
10/sin 51=12/sin Q
10(sin Q)/10=12(sin 51)/10
Q= 2nd function sin 0.9325751
Q=68 degrees
Q=180-68
Q=111 degrees
Answers
Answered by
Steve
Your alternate answer of 112° is not possible, since then 12 would not be the longest side. 68° is the only right choice.
rather than saying <u>2nd function sin</u> try saying sin^-1(.93) or arcsin(.93)
calculator buttons vary from model to model. We are talking math here, not buttons.
rather than saying <u>2nd function sin</u> try saying sin^-1(.93) or arcsin(.93)
calculator buttons vary from model to model. We are talking math here, not buttons.
Answered by
Amone
But when I tried to prove this it seems right. What I did,
Sin18(10)/(sin 51)=r
4=r
Cos Law:
4x4=(10x10)+(12x12)-2(12)(10)cos 18 degrees
16=100+144-228.23
16=15.96
Sin18(10)/(sin 51)=r
4=r
Cos Law:
4x4=(10x10)+(12x12)-2(12)(10)cos 18 degrees
16=100+144-228.23
16=15.96
Answered by
Amone
Isn't the Obtuse angle always opposite of the longest side? I'm confused, is there a big a misunderstanding on my side.
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