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In the diagram below AM = BM = CM, and $\angle BMC+\angle A = 210^\circ.$ Find $\angle B$ in degrees.

 

 Jul 31, 2021
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In the diagram below
\(AM = BM = CM\), and
\(\angle BMC+\angle A = 210^\circ\).
Find \(\angle B\) in degree

 

\(\begin{array}{|rcll|} \hline \angle BMC+\angle A &=& 210^\circ \\ \mathbf{\angle BMC} &=& \mathbf{210^\circ-\angle A} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \text{In }\triangle AMC \\ \hline (180^\circ-\angle BMC)+2*(\angle A) &=& 180^\circ \\ 2*(\angle A)&=& \angle BMC \quad | \quad \mathbf{\angle BMC=210^\circ-\angle A}\\ 2*(\angle A) &=& 210^\circ-\angle A \\ 3*(\angle A) &=& 210^\circ \\ \mathbf{\angle A} &=& \mathbf{70^\circ} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \text{In }\triangle BMC \\ \hline \angle BMC+2*(\angle B) &=& 180^\circ \quad | \quad \angle BMC = 2*(\angle A) \\ 2*(\angle A)+2*(\angle B) &=& 180^\circ \\ \angle A+\angle B &=& 90^\circ \\ \angle B &=& 90^\circ -\angle A \\ \angle B &=& 90^\circ -70^\circ \\ \mathbf{\angle B} &=& \mathbf{20^\circ} \\ \hline \end{array}\)

 

laugh

 Aug 1, 2021

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