Question

Find:-
1)angle B
2.angle EDC
3.abgleADE
4.angleAED
5.angle DEC
6.angle DEC
7.ANGLE ACB​

Answer 1

SOLUTION...

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Answer 2

Answer:

$\angle B = 55 \degree </p><p> \\ \angle EDC = 25 \degree </p><p> \\ \angle ADE = 55 \degree \\ \angle AED =70 \degree \\ \angle DEC = 110 \degree \\ \angle ACB = 70 \degree$

Step-by-step explanation:

In Δ DBC,

< CDB + < DBC + < DCB = 180°

100° + < DBC + 25° = 180°

< DBC = 180° - 125°

< DBC = 55° ; < B = 55°

< DCB = < CDE = 25° ( Vertically Opposite Angles )

< CDE = 25°

< CDB + < CDE + < EDA = 180° ( Linear Pair Theorem)

100° + 25° + < EDA = 180°

< EDA = 180° - 125°

< EDA = 55°

In ΔADE ,

< DAE + < ADE + < AED = 180°

55° + 55° + < AED = 180°

< AED = 180° - 110°

< AED = 70°

< AED + < DEC = 180° ( Linear Pair Theorem)

70° + < DEC = 180°

< DEC = 110°

In Δ DEC ,

< EDC + < DEC + < ECD = 180°

25° + 110° + < ECD = 180°

< ECD = 180° - 135°

• < ECD = 45°

< ACB = < ECD + < DCB

< ACB = 45° + 25°

< ACB = 70°