Suppose angle a=pi/6 radians and length of b=5

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Suppose angle a=pi/6 radians and length of b=5

Mathematics
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There are 3 angles in your triangle. One measures pi/3. Another angle is a right angle. You can calculate the measure of the third angle.
so angle A=30 degrees, and angle b=60 degrees

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What is the sum of the measures of the angles of a triangle?
the sum is 180 degrees.. got that down :) I'm just having trouble with finding the lengths.
Angle A does not measure 30 deg. Also, they want the answer in radians.
she has done similar problems before :) what have you tried for a and c ?
if you convert 30 degrees into radians you get pi/6
sin30=5/c, c*sin30=5,c=5/sin30, 5 divided by 1/2, = 5*2 which is 10
sin 30 = 5/c sure?? B is 60 degrees
oops that's the cosine
Yes, you are correct. I was thinking of angle B, which is the one they are asking for. 180 deg = pi rad The sum of the measures of the angles of a triangle is 180 deg = pi rad. m
Now let's look at finding the length of a.
yea, now try again
math, she is faster than you think ;)
:) cosine 30=5/c
we know that cosine is sqrt3/2
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good! so c = ... ?
c=10/sqrt(3)?
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yessss! now onto a
we can use tan30=a/5
so a =... ?
well we know that tan 30=sqrt(3)/3
sqrt 3/ 3 = 1/sqrt 3
5sqrt(3)/3?
yes, correct! 5 (sqrt 3)/3 or 5/ sqrt 3
one good way to verify, side opposite to 30 degree angle should come out to be half the hypotenuse :)
We know the measure of angle A and the length of the adjacent side, b. We want the length of the opposite side, a. The tangent function relates the lengths of the opposite and adjacent sides: \(\tan A = \dfrac{opp}{adj} \) \(\tan \dfrac{\pi}{6} = \dfrac{a}{5}\) \(a = 5 \tan \frac{\pi}{6} \) \(a = \dfrac{5 \sqrt{3}}{3} \)
\(\sin A = \dfrac{opp}{hyp}\) \(\sin \dfrac{\pi}{6} = \dfrac{\frac{5\sqrt{3}}{3}}{c}\) \(\dfrac{1}{2} = \dfrac{\frac{5\sqrt{3}}{3}}{c}\) \(c = \dfrac{10\sqrt{3}}{3}\)

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