Question

With the help of ruler and compass, it is not possible to construct an angle of :
\[
  A){60^o} \\
  B){15^o} \\
  C){38^o} \\
  D){135^o} \\
\]

Answer 1

Hint: Find angles that cannot be done by angle bisector using the compass.
We are first going to find the out whether we can be able to draw \[{60^o}\], as we know that we can easily draw \[{60^o}\] using the compass, as it is the angle which the compass first forms initially and next coming to \[{15^o}\], we can even draw this angle using compass, as when we bisect the angle \[{60^o}\] using compass, we get the angle ${30^o}$, then when further bisect this angle, the resultant angle which we are going to obtain is \[{15^o}\]. The similar way, we are going to find the bisected value gives the resultant angles.

Complete step by step answer:
Consider \[A){60^o}\], this can be easily drawn with help of the compass, as the first arc formed after bisecting at equal lengths gives an initial angle of \[{60^o}\].
Now, next we consider \[B){15^o}\], we can get this angle, as well, when we initially get the angle of \[{60^o}\], when we perform angular bisector using compass, we get the angle ${30^o}$, when further bisect the angle we get the required angle \[{15^o}\].
Now, consider \[C){38^o}\], this angle cannot be drawn with the help of bisecting any of the angles.
Now, consider \[D){135^o}\], this angle lies between ${90^o}$ and \[{180^o}\] and can be done by doing an angle bisector between those and get the angle.

So, the correct answer is Option C.

Note: We have considered the two points according to the constants present in the given arguments and they are not always present around the present considered point as the locus of the complex number can be representing any other figure from the given options.