Question

5. In the given figure, side BC of a triangle ABC
has been produced to D and CE II BA. If
<ABC = 65°, <ACE = 55°, find <BAC, <ACB
and <ECD.​

Answer 1

Answer:

CE II BA

then

angle BAC = angle ACE ( alternate angles)

BAC = 55⁰

ABC + BAC + ACB = 180⁰

65⁰+55⁰+ACB =180⁰

120⁰ + ACB = 180⁰

ACB = 180⁰-120⁰

ACB = 60⁰

60⁰+55⁰+ECD = 180⁰

115⁰+ ECD = 180⁰

ECD = 180⁰-115⁰

ECD = 65⁰

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Answer 2

$\huge\frak{Answer:-}$

Given:-

• $CE || BA$,
• $\angle ABC = 65°$
• $\angle ACE = 55°$

To Find: $\angle BAC$,$\angle ACB$ & $\angle ECD$.

_______________...

$\underline{\tt{\angle BAC:-}}$

$\implies \angle BAC = \angle ACE$

$\bf{(Corresponding \ Angles)}$

Or,

$\angle BAC = 55°$

$\underline{\tt{ACB:-}}$

$\angle ABC + \angle BAC + \angle ACB = 180°$

$\bf{(Angle \ sum \ prop.)}$

$\implies 65° + 55° + \angle BAC = 180°$

$\implies 120° + \angle BAC = 180°$

$\implies \angle BAC = 180° - 120°$

$\implies \angle BAC = 60°$

$\underline{\tt{\angle ECD:-}}$

$\angle BAC + \angle ABC = \angle ACD$

$\bf{(Sum \ of \ int. \ angles \ is \ equal \ to \ the \ ext. \ angle)}$

$\implies 55° + 65° = \angle ACE + \angle ECD$

$\implies 55° + 65° = 55° + \angle ECD$

$\implies \angle ECD = 65°$