Math, asked by mahatonakul27, 9 months ago

In the given figure <ABC = 50°,
<BAC = 25°, CDE = 60°, find <AED.

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Answered by Anu1612

∠ACB=180°-25°-50°

=180°-75°

=105°

∠ECD=180°-105° (since BCD is a straight line)

= 75°

∠AED=60°+75° (since sum of two interior angles = 1 exterior angle)

=135°

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Answered by Sambhavs

Answer:

$\huge\blue{ANSWER}$

WE KNOW THAT SUM OF 3 ANGLES OF A TRIANGLE IS =180⁰

NOW CONSIDERING ∆ABC

ANGLE ARE AT VERTICES A,B,C

NOW we know that

Angle A = 25⁰

Angle B = 50⁰

Then Angle C = 180⁰- 25⁰+50⁰ (by Angle sum property)

SO ANGLE C = 105⁰

NOW ANGLE D = 180⁰-50⁰ = 130⁰(ANGLES ON A STRAIGHT LINE)

Now Angle ${E}$

ANGLE E = 180⁰-50⁰=130⁰(A) BECAUSE ANGLE ON A STRAIGHT LINE

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In the given figure <ABC = 50°,<BAC = 25°, CDE = 60°, find <AED.​

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In the given figure <ABC = 50°,
<BAC = 25°, CDE = 60°, find <AED.