give steps for cons and justification for 105 and 135 degrees

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Please find below the solution to the asked query:

We follow these steps to construct angles of 105$°$ and 135$°$ :

Step 1: Draw line PX and take any point " Q " as center and any radius ( Less than half of PQ ) and draw a semicircle that intersect our line PX at " A "  and " B " .

Step 2 : With same radius and center " A " draw an arc that intersect our semicircle at " C " , Again with same radius and center " C " draw an arc that intersect our semicircle at " D " .

Step 3 : With same radius and center " C " and " D " draw arcs and these arcs intersect  at " E " . Join QE and line QE intersect our semicircle at " F " .

Step 4 : With same radius and center " D " and " F " draw arcs and these arcs intersect  at " R " . Join QR , Here $\angle$ PQR  = 105$°$.

Justification : $\angle$ PQE = 90$°$ and arc " D "  denotes 120$°$ with center " Q " , so $\angle$ PQD = 120 $°$, $\angle \mathrm{PQR} = \frac{\angle \mathrm{PQE}\hspace{0.17em}+ \angle \mathrm{PQD}}{2} = \frac{90°+ 120°}{2}= \frac{210°}{2} = 105°$

Step 5 : With same radius and center " B " and " D " draw arcs and these arcs intersect  at " G " . Join QG and line QG intersect our semicircle at " H " .

Step 6 : With same radius and center " D " and " H " draw arcs and these arcs intersect  at " S " . Join QS , Here $\angle$ PQS  = 135$°$.

Justification : $\angle$ PQG = 150$°$ and arc " D "  denotes 120$°$ with center " Q " , so $\angle$ PQD = 120 $°$, $\angle \mathrm{PQS} = \frac{\angle \mathrm{PQG}\hspace{0.17em}+ \angle \mathrm{PQD}}{2} = \frac{150°+ 120°}{2}= \frac{270°}{2} = 135°$


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