# In triangleABC, let angleA=60, side b=10 and side c=7. Find side a?what do you have to draw?

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### 2 Answers

You need to use the law of cosines to find a such that:

`a^2 = b^2 + c^2 - 2bc*cos A`

You need to substitute 10 for b,7 for c and 60^o for A such that:

`a^2 = 100 + 49 - 140 cos 60^o`

You need to substitute `1/2` for `cos 60^o` such that:

`a^2 = 149 - 70 =gt a^2 = 79`

`a = +-sqrt 79`

You need to keep the positive value for the length a, thus `a = sqrt 79.`

**Hence, evaluating the length of side a yields `a = sqrt 79` .**

Draw a height from point C to side c called CD.

The height CD=b*sinA=bsinA

The length AD=b*cosA=bcosA

since AD+DB=c

so DB=c-AD=c-bcosA

The side a=sqrt(CD^2+DB^2)=sqrt((bsinA)^2+(c-bcosA)^2)

=sqrt(b^2(sinA)^2+c^2-2bc*cosA+b^2(cosA)2)

=sqrt(b^2+c^2+abc*cosA)

=sqrt(10^2+7^2-2*10*7*cos60)=sqrt(79)

about 8.888