+0  
 
+1
179
1
avatar+598 

In right triangle $ABC$, we have $AB = 10$, $BC = 24$, and $\angle ABC = 90^\circ$. If $M$ is on $\overline{AC}$ such that $\overline{BM}$ is a median of $\triangle ABC$, then what is $\cos \angle ABM$?

michaelcai  Nov 15, 2017
Sort: 

1+0 Answers

 #1
avatar
+2

cos(ABM) = ?

 

\(AM=MC=BM=r(\text{circle })\)

 

\(\tan{ACB}=\frac{10}{24} \)

 

\(\frac{\sin(ACB)}{r}=\frac{\sin(90^{\circ}-ABM)}{r}\\ \sin(ACB)=\cos(ABM)\\ \cos(ABM)=\sin(\arctan(\frac{10}{24}))=0.384615384615...\\ ABM=\arccos(\frac{5}{13})=67.380135051983^{\circ } \)

 

laugh heureka

Guest Nov 15, 2017
edited by Guest  Nov 15, 2017

19 Online Users

New Privacy Policy (May 2018)
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  Privacy Policy