# What is the length of the third side of a triangle if two sides are of length 3 and 6 and the angle between them is 30 degrees. Also, what are the other 2 angles.

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Given two side lengths of a triangle, 3 and 6, and the angle between them of `30^@ ` , find the missing length and the other two angles:

Method I: Recognize this as a 30-60-90 right triangle, so the missing side is `3sqrt(3) ` and the missing angles are `60^@,90^@ ` .

Method II: Since we have two sides and the included angle, use the Law of Cosines. With c the missing side and C the given angle we use:

`c^2=a^2+b^2-2abcosC `

`c^2=3^2+6^2-2(3)(6)cos(30^@) `

`c^2=45-36(1/2) ` Recognize `cos30^@ ` from the unit circle, or use a calculator in degree mode.

`c^2=27 ==> c=sqrt(27)=3sqrt(3) `

Then you can use the Law of Sines to determine one of the missing angles:

`(3sqrt(3))/(sin(30^@))=3/(sinA) `

`3sqrt(3)sinA=(3sqrt(3))/2 `

`sinA=1/2==>A=60^@ ` (** `sin^(-1)(1/2)=60^@ ` **)

Since the angle sum of a triangle is 180, the third angle has measure 90.

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I am sorry, I reversed the sin and cos of 30 degrees:

side length: `c^2=45-18sqrt(3)~~3.7179 `

angles: `3.7179/(sin(30))=3/(sinA) `

`3.7179sinA=1.5 `

`sinA~~.4035==>A~~23.8^@ ` and the third angle is 180-30-23.8 or `B~~126.2^@ `

The first method is clearly incorrect.

The triangle in the problem has two sides with length 3 and 6 and the angle between them is 30 degrees.

To determine the length of the third side use the Law of cosines. If the angles of a triangle are A, B and C and the length of the opposite sides are a, b and c, the law of cosines gives c^2 = a^2 + b^2 - 2*a*b*cos C.

In the problem, a = 3, b = 6 and C = 30

c^2 = 3^2 + 6^2 - 2*3*6*cos 30

=> `c ~~ 3.7179`

The other angles can be determined using the law of sines: `a/sin A = b/sin B = c/sin C`

`3.7179/sin 30 = 3/sin A`

=> `A = sin^-1(3/(2*3.7179)) ~~ 23.79`

`B = sin^-1(6/(2*3.7179)) ~~ 126.2`

The third side of the triangle is of length 3.7179. The other two angles are 23.79 and 126.2 degrees.