Question

In figure angle COD = 90° ,angle BOE =72° and angle AOB =180° ( straight angle) Find measures of the following angles: angle AOC ,angle BOD ,angle BOC ,angle AOE​

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

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1) Draw a line segments AB = 6.4 cm.

2) With A s a centre and a radius equal to more than half of AB, draw two arcs,one above AB and other below AB.

3),With B as centre and the same radius, draw two arcs, cutting the previous drawn arcs at point C and D respectively.

4)Join CD, intersecting AB at point O.

Then, CD is the required perpendicular bisectors of AB at the point O.

On measuring, we find that

OA = 3.2 cm and OB = 3.2 cm

Also, ∠AOC = ∠BOC = 90°

$\bf\underline{\underline{\blue{Justification:-}}}$

Join AC, AD, BC and BD.

In ∆CAD and ∆CBD, we have

AC = BC (arcs of equal radii)

AD = BD (arcs of equal radii)

CD = CD (common)

∴∆CAD ≅ ∆CBD (S.S.S-criterian)

∴∠ACO = ∠BCO (c.p.c.t)

Now, in ∆AOC and ∆BOC, we have

AC = BC (arcs of equal radii)

∠ACO = ∠BCO (proved above)

CO = CO (common)

∴ ∆AOC ≅ ∆BCO (S.A.S-criterian)

Hence, AO = BO and ∠AOC = ∠BOC

But, ∠AOC + ∠BOC = 180° (linear pair axiom)

∴ ∠AOC = ∠BOC = 90°

Hence, COD is the perpendicular bisector of ∠AOB.