Two sides of a triangle are 3+1 and 3−1 and the included angle is 60∘. Find other angles My Attempt Let a=3+1, b=3−1 and C=60 c2=a2+b2−2.a.bcos⁡C =(3+1)2+(3−1)2−2(3+1)(3−1).12=8−2=6 ⟹c=6=23 asin⁡A=csin⁡C⟹sin⁡A=3+12332=3+122 A=75∘&amp;B=45∘ But the solution given in my reference is 105∘ and 15∘, what is going wrong with my attempt ?

Question

Two sides of a triangle are 3+1 and 3−1 and the included angle is 60∘. Find other angles My Attempt Let a=3+1, b=3−1 and C=60 c2=a2+b2−2.a.bcos⁡C =(3+1)2+(3−1)2−2(3+1)(3−1).12=8−2=6 ⟹c=6=23 asin⁡A=csin⁡C⟹sin⁡A=3+12332=3+122 A=75∘&amp;amp;B=45∘ But the solution given in my reference is 105∘ and 15∘, what is going wrong with my attempt ?

Joy Compton · Accepted Answer

Because sin⁡(α)=sin⁡(180−α). Hence, sin⁡(75∘)=sin⁡(105∘). Therefore you should compute the value of sin⁡B instead to specify one of them. Probably, when you computing the value of angle B you will get 165∘  and  15∘, but as the sum of angles is equal to 180∘,15∘ will be accepted.

Two sides of a triangle are \sqrt3+1 \text{ and }

Answered question

Answer & Explanation

New Questions in Precalculus