HELP MEDAL!!
Solve the triangle. A = 21°, C = 105°, c = 5
@robtobey
please help
B = 180-(21+105) = 54 Use the Law of Sines to calculate the other 2 sides. http://www.mathsisfun.com/algebra/trig-sine-law.html
B = 54°, a ≈ 13.5, b ≈ 4.2 right? @robtobey
?
@cwrw238
Recalculate a. I got a = \[a=\frac{10 \sqrt{2} \sin (21 {}^{\circ})}{1+\sqrt{3}}=1.85501 \]
b = 4.18778
nope youre definetly wrong, none of my choices
@SithsAndGiggles please help
uhmmm?
|dw:1408121921631:dw| \[180=B+21+105~~\iff~~B=54^\circ\] Law of sines gives \[\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}~~\iff~~\frac{\sin21^\circ}{a}=\frac{\sin54^\circ}{b}=\frac{\sin 105^\circ}{5}\] For \(a\) and \(b\) you get \[a=\frac{5\sin21^\circ}{\sin105^\circ}\approx1.855\] \[b=\frac{5\sin54^\circ}{\sin105^\circ}\approx4.188\] Getting the same as @robtobey...
whatttt. these are my choices
oops wrong one
ohh!!!!
It's the first one. 1.855 gets rounded up to 1.9, 4.188 gets rounded to 4.2
yes! thank you so much!!!
yw
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