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
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 2nd function sin try saying sin^-1(.93) or arcsin(.93)
calculator buttons vary from model to model. We are talking math here, not buttons.
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
Isn't the Obtuse angle always opposite of the longest side? I'm confused, is there a big a misunderstanding on my side.
To solve for Q in triangle PQR, we can use the Law of Sines:
sine(Q) / 12 = sine(51) / 10
To find Q, we can isolate the sine(Q) term:
sine(Q) = (12 * sine(51)) / 10
Divide both sides by 12:
sine(Q) / 12 = sine(51) / 10
Now we can take the inverse sine (sin^(-1)) of both sides to solve for Q:
Q = sin^(-1)((12 * sine(51)) / 10)
Using a calculator, we can find:
Q ≈ 68 degrees
Since Q is an angle in an obtuse triangle, it cannot be greater than 90 degrees. Therefore, we can deduce that Q is actually the supplement of 68 degrees:
180 - 68 = 112 degrees
Therefore, Q is approximately 112 degrees (to the nearest degree).