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Question
mn is tangent to ⊙k. what is m∠p? m∠p =
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pythagorean theorem (100)
lesson: tangent to a circle
Step1: Recall tangent - radius property
A tangent to a circle is perpendicular to the radius at the point of tangency. So, $\angle KMN = 90^{\circ}$.
Step2: Note that $\triangle KMN$ and $\triangle KPN$
Since $KP = KM$ (radii of the same circle), $\triangle KMN$ and $\triangle KPN$ are related. Also, $\angle MKN = 180^{\circ}- 2\angle P$ (central - inscribed angle relationship for the arc $MN$). In $\triangle KMN$, we know that $\angle N = 27^{\circ}$ and $\angle KMN=90^{\circ}$. Then, $\angle MKN=180^{\circ}-\angle N - \angle KMN$.
$\angle MKN=180^{\circ}-27^{\circ}-90^{\circ}=63^{\circ}$.
Step3: Use the angle - relationship for inscribed and central angles
We know that the central angle is twice the inscribed angle subtended by the same arc. Let $\angle P$ be the inscribed angle and $\angle MKN$ be the central angle subtended by arc $MN$. Also, in $\triangle KPN$, since $KP = KN$ (radii), we can use the fact that $\angle MKN = 2\angle P$.
If $\angle MKN = 63^{\circ}$, then $\angle P=\frac{1}{2}\angle MKN$.
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