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Solve the triangle. A = 33°, a = 20, b = 13

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asked Jul 31, 2014 in TRIGONOMETRY by Tdog79 Pupil

1 Answer

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Best answer

angle A = 33o and sides a = 20, b = 13.

Lets find remaining angles are B and C and the remaining side c.

Law of sines : a/sinA = b/sinB ⇒ sinB = (b*sinA)/a

sinB = (13*sin33o)/20 = (13*0.5446)/20 ≅ 0.354

B ≅ sin(0.354) ≅ 20.733

Since sum of angles in a triangle = 180.

A+ B +C = 180.

Angle C = 180 - (A + B) = 180 - 53.733 = 126.266.

Law of sines :  b/sinB = c/sinC ⇒ c = b*sinC/sinB

c = (13)*sin(126.266)/sin(20.733) = (13)*(0.806)/(0.354) = 29.598887 = 30.

answered Jul 31, 2014 by casacop Expert
Thank you C:
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