Question

construct an angle of 45°at the initial point of a given ray and justify the construction​

Answer 1

$\huge\star{\pink{\underline{\tt{Explanation:}}}}$

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$\large{\purple{\tt{STEPS-OF-CONSTRUCTION:-}}}$

$\leadsto$Draw a ray OA

$\leadsto$ Take O as a centre with any radius, draw an arc DCB is that cuts OA at B.

$\leadsto$With B as a centre with the same radius, mark a point C on the arc DCB.

$\leadsto$ With C as a centre and the same radius, mark a point D on the arc DCB.

$\leadsto$ Take C and D as centre, draw two arcs which intersect each other with the same radius at P.

$\leadsto$ Finally, the ray OP is joined which makes an angle 90° with OP is formed.

$\leadsto$ Take B and Q as centre draw the perpendicular bisector which intersects at the point R

$\leadsto$ Draw a line that joins the point O and R

$\leadsto$ So, the angle formed ∠ROA = 45°

$\huge\star{\blue{\tt{\underline{Justification:-}}}}$

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From the construction,

∠POA = 90°

From the perpendicular bisector from the point B and Q, which divides the ∠POA into two halves. Therefore,

∠ROA = 1/2 ∠POA

∠ROA = 1/2 × 90° = 45°

$\large{\red{\bf{\underline{Hence,Solved!}}}}$

Answer 2

heey buddy

hope it's useful