Question

plz answer this question​

Answer 1

Given :-

• ∠BAC = 25°
• ∠CBA = 45°
• ∠EDC = 40°

To find :-

• ∠ACD = ?
• ∠AED = ?

How to find ?

• First we have to find the angle ACB so, because that we can find angle ACD by linear pair. After all we can find Angle CED by angle sum property of triangle and finally we can find angle AED by linear angle property. Hence we can find both angles.

Property's we will used :-

• Angle sum property of triangle :- It states that sum of all angle of triangle is 180°

• Linear pair :- It states that sum of two angles of same line is 180°

So, in △ ABC. (By angle sum property)

$\mathtt{ \implies∠BAC + ∠CBA + \angle \: ACB = 180 \degree }$

$\mathtt{ \implies25 \degree+ 45 \degree + \angle \: ACB = 180 \degree}$

$\mathtt{ \implies70 \degree + \angle \: ACB = 180 \degree}$

$\mathtt{ \implies \angle \: ACB = 180 \degree -70 \degree }$

$\large \color{green}\mathtt{ \implies \angle \: ACB = 110 \degree}$

So, ∠ACD. (By Linear pair )

$\mathtt{ \implies \angle \: ACB + ∠ACD = 180 \degree}$

$\mathtt{ \implies 110 \degree + ∠ACD = 180 \degree}$

$\mathtt{ \implies ∠ACD = 180 \degree - 110 \degree }$

$\boxed{\color{red}\LARGE \mathtt{ \implies ∠ACD = 70\degree}}$

In △CED. (Angle sum property)

$\mathtt{\implies \angle ACD + ∠EDC+\angle CED = 180°}$

$\mathtt{\implies 70 \degree+40 \degree + \angle CED = 180°}$

$\mathtt{\implies 110 \degree + \angle CED = 180°}$

$\mathtt{\implies \angle CED = 180° - 110 \degree }$

$\large \color{green}\mathtt{\implies \angle CED = 70\degree }$

So, ∠AED. (By linear pair )

$\mathtt{\implies {\angle CED + ∠AED = 180°}}$

$\mathtt{\implies {70°+ ∠AED = 180°}}$

$\mathtt{\implies {∠AED = 180°- 70°}}$

$\boxed{\LARGE{\color{red}\mathtt{\implies {∠AED = 110°}}}}$

So , ∠AED & ∠ACD are 110° and 70° respectively.