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The given measurements may or may not determine a triangle. If not, then state that no triangle is formed. If a triangle is formed, then use the Law of Sines to solve the triangle, if it is possible, or state that the Law of Sines cannot be used.
C = 38°, a = 19, c = 10
 one year ago
 one year ago
will give medal to best answer! The given measurements may or may not determine a triangle. If not, then state that no triangle is formed. If a triangle is formed, then use the Law of Sines to solve the triangle, if it is possible, or state that the Law of Sines cannot be used. C = 38°, a = 19, c = 10
 one year ago
 one year ago

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HeroBest ResponseYou've already chosen the best response.1
Did you try simply using Law of Sines to figure it out?
 one year ago

meg1334Best ResponseYou've already chosen the best response.0
yes but i didnt understand
 one year ago

HeroBest ResponseYou've already chosen the best response.1
What did you come up with?
 one year ago

meg1334Best ResponseYou've already chosen the best response.0
i got sin A = 67 but that isnt in my answer choices No triangle is formed. A = 58.6°, B = 83.4°, b ≈ 15.6 A = 83.4°, B = 58.6°, b ≈ 15.6 The triangle cannot be solved with the Law of Sines.
 one year ago

HeroBest ResponseYou've already chosen the best response.1
I don't understand how you got that. Please show me the steps you took to get A = 67
 one year ago

meg1334Best ResponseYou've already chosen the best response.0
i used c/ sin c = a/ sin a 10/sin 37 = 19/sin a and then i divided 19 sin 38 over sin 10 and i got 67.36
 one year ago

HeroBest ResponseYou've already chosen the best response.1
where did you get the 37 from?
 one year ago

meg1334Best ResponseYou've already chosen the best response.0
I meant 38, i used 38 in my claculations and still got 67
 one year ago

HeroBest ResponseYou've already chosen the best response.1
Are you in degree mode or radian mode?
 one year ago

meg1334Best ResponseYou've already chosen the best response.0
i dont know i just typed it in and got 67, cant you check it?
 one year ago

HeroBest ResponseYou've already chosen the best response.1
You must make sure you're in degree mode. That's very important.
 one year ago

HeroBest ResponseYou've already chosen the best response.1
But either way it goes, you're not doing it correctly.
 one year ago

meg1334Best ResponseYou've already chosen the best response.0
oh really, i hadnt noticed that im doing it wrong. I need help thats why im on here obviously
 one year ago

HeroBest ResponseYou've already chosen the best response.1
\[\sin A = \frac{19\sin(38^{\circ})}{10} \\\sin A = \]
 one year ago

HeroBest ResponseYou've already chosen the best response.1
Compute the right hand side again and let me know what you get. Make sure you're in degree mode.
 one year ago

HeroBest ResponseYou've already chosen the best response.1
Okay now \[\sin A = 1.169756803\] So how do we find A?
 one year ago

meg1334Best ResponseYou've already chosen the best response.0
a is 19, sin A is 1.16, what do you mean find A?
 one year ago

HeroBest ResponseYou've already chosen the best response.1
I'm not talking about a = 19. That's a side length
 one year ago

HeroBest ResponseYou've already chosen the best response.1
\[\sin (A^{\circ}) = 1.169756803\] I'm talking about Angle A. Find that.
 one year ago

HeroBest ResponseYou've already chosen the best response.1
You take the inverse sine of both sides to get: \[\sin^{1}(A^{\circ}) = \sin^{1}({1.169756803}) \\A^{\circ} = \sin^{1}({1.169756803}) \]
 one year ago

HeroBest ResponseYou've already chosen the best response.1
So compute \[\sin^{1}({1.169756803})\]
 one year ago

HeroBest ResponseYou've already chosen the best response.1
Let me know what you get.
 one year ago

HeroBest ResponseYou've already chosen the best response.1
That's what it should say because no such angle exists.
 one year ago

HeroBest ResponseYou've already chosen the best response.1
And if the angle doesn't exist, then the triangle can't possibly exist.
 one year ago

HeroBest ResponseYou've already chosen the best response.1
\[1< \sin(\theta) < 1\] means that the value of sine can only be between  or + one. So there's no such thing as \(\sin(\theta) = 1.17\)
 one year ago
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