Find the value of angle $x$ in the polygon below

euclidean-geometrygeometryplane-geometry

The graph shows two regular polygons. If $O$ is the center of one of them, calculate $x$ (Answer: $82,5^{\circ})$

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I found some Angles and tried to use the $\triangle DTS$ but I couldn't demonstrate that it is a right triangle.

$\angle NML = 135^{\circ} \\
\angle LDP = 120^{\circ} \\
\angle SDO = 60^{\circ}\\
\angle LDS = \dfrac{135^{\circ}-120^{\circ}}{2} = 7,5^{\circ}$

enter image description here

Best Answer

You've already done most of the work. The sum of angles along a straight line being $180^{\circ}$ means that

$$\measuredangle TML = \measuredangle TLM = 180^{\circ} - 135^{\circ} = 45^{\circ}$$

From the sum of angles in $\triangle LTM$, we then have

$$\measuredangle LTM = 180^{\circ} - 45^{\circ} - 45^{\circ} = 90^{\circ}$$

Using your result of $\measuredangle LDS = 7.5^{\circ}$ and the sum of angles in $\triangle DTS$, we get that

$$x = \measuredangle TSD = 180^{\circ} - 7.5^{\circ} - 90^{\circ} = 82.5^{\circ}$$