Question

In the given figure,triangle A B C,A B is parallel to D E,<B A C=35 degree and <C D E=53 degree,then find <A C D

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Answer 1

Step-by-step explanation:

<AED will be 35 degree according to property

Now, DEC forms a triangle, so, <DCE= 92 degrees

Now, <ACD + <DCE = 180 degrees

Thus, <ACD + 92 degrees = 180 degrees

At last, <ACD = 88 degrees

Answer 2

Answer:

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.

$\huge \underline{ \underline \purple{ \mathtt{GIVEN }} }:$

$\hookrightarrow \large{AB \parallel DE}$

$\hookrightarrow \angle \: bac = 35 \degree$

$\hookrightarrow \: \angle \: cde \: = 53 \degree$

$\hookrightarrow \large{ \angle \: bac \: = \angle \: ced} \:$

$\bull \: \: \: \: \footnotesize{ since \: they \: are} \: \red{alternate \: interior \: angles}$

$\therefore \huge \boxed{\red{ \angle \: ced \: = 35 \degree}}$

Now in ∆ CDE,

sum of all angles = 180°

$35 + 53 + \angle dce \: = 180 \degree$

$\angle \: dce = 92 \degree$

Now in line AE ,

$\hookrightarrow \angle \: acd \: + 92 = 180 \degree$

$\bull \: \: \: \: \: \: \: \: \footnotesize{ since \: they \: are \: } \footnotesize\red{linear \: pair}$

$\large\hookrightarrow \huge \red{ \boxed{\angle \: acd \: = 88 \degree}}$

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