#1**+1 **

**In triangle ABC, AB = BC = 25 and AC = 40. What is \(\sin( \angle ACB)\) ?**

**Heron's formula states that the area of a triangle whose sides have lengths \(a \), \(b\), and \(c\) is**

**\(\qquad A = \sqrt{s(s-a)(s-b)(s-c)}\), **

**where s is the semi-perimeter of the triangle; that is, **

**\(\qquad s=\dfrac{a+b+c}{2}\)**

\(\begin{array}{rcll} \text{Let $BC=a=25$} \\ \text{Let $AC=b=40$} \\ \text{Let $AB=c=25$} \\ \end{array}\qquad \begin{array}{|rcll|} \hline \mathbf{ s } &=& \mathbf{ \dfrac{a+b+c}{2} } \\ s &=& \dfrac{25+40+25}{2} \\ \mathbf{ s } &=& \mathbf{45 } \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \mathbf{ A } &=& \mathbf{\sqrt{s(s-a)(s-b)(s-c)} } \\ A &=& \sqrt{45(45-25)(45-40)(45-25)} \\ A &=& \sqrt{45*20*5*20} \\ A &=& \sqrt{15^2*20^2} \\ \mathbf{ A } &=& \mathbf{15 * 20} \\ \hline \end{array}\)

**Formula: **\(2A = ab\sin(\angle ACB)\)

\(\begin{array}{|rcll|} \hline 2A &=& ab\sin(\angle ACB) \quad | \quad \mathbf{ A = 15 * 20},\ a=25,\ b=40 \\ 2*15 * 20 &=& 25*40 \sin(\angle ACB) \\ 15 * 40 &=& 25*40 \sin(\angle ACB) \quad | \quad : 40 \\ 15 &=& 25 \sin(\angle ACB) \\ \sin(\angle ACB) &=& \dfrac{15}{25} \\ \mathbf{ \sin(\angle ACB) } &=& \mathbf{\dfrac{3}{5}} \\ \mathbf{ \sin(\angle ACB) } &=& \mathbf{0.6} \\ \hline \end{array} \)

heureka Aug 5, 2020