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# help

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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 an altitude of $\triangle ABC$, then what is $\cos \angle ABM$?

Guest Jun 28, 2018

#1
+20013
+2

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 an altitude of $\triangle ABC$,
then what is $\cos \angle ABM$?

$$\begin{array}{|rcll|} \hline \angle \text{ABM} &=& 90^{\circ}-\angle \text{CAB} \\ \cos(\angle\text{ABM}) &=& \cos(90^{\circ}-\angle \text{CAB}) \\ \cos(\angle \text{ABM}) &=& \sin(\angle \text{CAB}) \quad & | \quad \sin(\angle \text{CAB}) = \dfrac{BC}{\sqrt{AB^2+BC^2} } \\ \cos(\angle \text{ABM}) &=& \dfrac{BC}{\sqrt{AB^2+BC^2} } \\\\ \cos(\angle \text{ABM}) &=& \dfrac{24}{\sqrt{10^2+24^2} } \\\\ \cos(\angle \text{ABM}) &=& \dfrac{24}{\sqrt{676} } \\\\ \cos(\angle \text{ABM}) &=& \dfrac{24}{ 26 } \\\\ \mathbf{ \cos(\angle \text{ABM}) } & \mathbf{=} & \mathbf{ \dfrac{12}{ 13 } } \\ \hline \end{array}$$

heureka  Jun 28, 2018
#1
+20013
+2

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 an altitude of $\triangle ABC$,
then what is $\cos \angle ABM$?

$$\begin{array}{|rcll|} \hline \angle \text{ABM} &=& 90^{\circ}-\angle \text{CAB} \\ \cos(\angle\text{ABM}) &=& \cos(90^{\circ}-\angle \text{CAB}) \\ \cos(\angle \text{ABM}) &=& \sin(\angle \text{CAB}) \quad & | \quad \sin(\angle \text{CAB}) = \dfrac{BC}{\sqrt{AB^2+BC^2} } \\ \cos(\angle \text{ABM}) &=& \dfrac{BC}{\sqrt{AB^2+BC^2} } \\\\ \cos(\angle \text{ABM}) &=& \dfrac{24}{\sqrt{10^2+24^2} } \\\\ \cos(\angle \text{ABM}) &=& \dfrac{24}{\sqrt{676} } \\\\ \cos(\angle \text{ABM}) &=& \dfrac{24}{ 26 } \\\\ \mathbf{ \cos(\angle \text{ABM}) } & \mathbf{=} & \mathbf{ \dfrac{12}{ 13 } } \\ \hline \end{array}$$

heureka  Jun 28, 2018