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

 Aug 5, 2020
 #1
avatar+25543 
+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: \(2A = ab\sin(\angle ACB)\)

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

 

 

laugh

 Aug 5, 2020
 #2
avatar+1326 
+1

In triangle ABC, AB = BC = 25 and AC = 40. What is sin∠ACB?

 

Let M be a midpoint of AC

 

BM = sqrt( BC2 - MC2 )

 

sin∠ACB = BM / BC

 Aug 5, 2020

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