Question

please solve this problem please with solutions

Answer 1

Here your both answer
If it is wrong ask me
Thanks

Answer 2

10.

it is very easy it is a pentagon and sum of all angle of a pentagon is 540°

another way is

we have to find ∠A + ∠B+ ∠C +∠D + ∠E

this can be written as, ∠A + ∠B+ ∠C +∠D + ∠E = ∠A + ∠B+ ∠BCE + ∠DCE + ∠D + ∠AEC + ∠DEC  [∠C = ∠BCE + ∠DCE & ∠E= ∠AEC + ∠DEC ]

= ∠A + ∠B+ ∠BCE + ∠AEC + ∠DCE + ∠DEC + ∠D

= 360° + 180° [∠A + ∠B+ ∠BCE + ∠AEC = 360° (angles of a quadrilateral)]  & [∠DCE + ∠DEC + ∠D = 180° (angles of a triangle)]

=540°

Hence ∠A + ∠B+ ∠C +∠D + ∠E = 540°

                                                Part 2

a)

if PR = RQ

then ∠RPQ = ∠RQP

Hence ∠RQP = 80°

Now in triangle PQR

∠RPQ + ∠RQP + ∠PRQ = 180° [sum ofall angle of triangle]

80° + 80° ∠PRQ = 180°

∠PRQ = 180° - 160°  = 20°

b)

a + ∠RQP = 180° [linear pair has a sum of 180°]

a = 180° - ∠RQP

a = 180° - 80°

a=100°

again

b + ∠PRQ = 180° [linear pair has a sum of 180°]

b = 180° - ∠PRQ

b = 180° - 20°

b = 160°

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