Question

please explaine it ?.............​

Answer 1

hii buddy here is answer .

at last I used ex angle property .

Answer 2

${ \dagger{\red{ \bold{ \: \: \: given}}}}$



⛦ AC ⊥ BD



Then,


∠ACB = ∠ ACD = 90°






${ \dagger{ \red{ \bold{ \: \: \: to \: find}}}}$



∠ AED = ???





${ \dagger{ \red{ \bold{ \: \: \: solution}}}}$





In Δ ABC ,



☞ ∠ BAC = x


☞ ∠ ABC = ( 4x - 10° )


☞ ∠ ACB = 90°




${ \sf{using \: \: { \underline{ \green{angle \: sum \: property \: of \: triangle}}}}}$




${ \boxed{ \boxed{ \blue{ \sf{sum \: of \: all \: 3 \: angles = 180}}}}}$



∠ ACB + ∠ ABC + ∠ BAC = 180°




➜ 90° + ( 4x - 10° ) + x = 180°



➜ 90° - 10° + 4x + x = 180°



➜ 80° + 5x = 180°



➜ 5x = 180° - 80°




➜ 5x = 100°



➜ x = 20°





In Δ CDE,



☞ ∠ ACD = ∠ ECD = 90°



☞ ∠ CDE = ( 3x - 6 ) = ( 3*20 - 6 )°

= 60° - 6° = 54°





${ \sf{ using \: \: { \underline{ \green{exterior \: angle \: property \: of \: triangle}}}}}$



An exterior angle of a triangle is equal to the sum of the opposite interior angles.



∠ ECD + ∠ CDE = ∠ AED



➜ 90° + 54° = 6y


➜ 6y = 144°


➜ y = 24°



thus,



⛦ ∠ AED = 6y = 144°