Question

in a given figure side BC of the triangle ABC is produced to D. if angle ACD = 128° and angle ABC = 43°. find angle BHC and angle ACB​

Answer 1

$\star\;{\underline{\rm{\red{\;\;GivEn:-\;\;}}}}$

• ∠ ACD = 128°

• ∠ ABC = 43°

$\star\;{\underline{\rm{\blue{\;\;To\;Find:-\;\;}}}}$

• ∠ BAC and ∠ ACB

$\star\;{\underline{\rm{\green{\;\;SoluTion:-\;\;}}}}$

In ∆ ABC -

• ∠ ACD = 128°

• ∠ ABC = 43°

⋆ Reference of image is shown in "Attachment".♡

$:\implies$ ∠ ACD = ∠ BAC + ∠ ABC [ ∵ An exterior angle of a triangle is ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀equal to the sum of the opposite interior angles.]

$:\implies$ 128° = ∠ BAC + 43°

$:\implies$ 128° =∠ BAC + 43°

$:\implies$ 128° - 43° = ∠ BAC

$:\implies{\underline{\boxed{\bf{\pink{ \angle BAC = 85^\circ}}}}}$ ★

━━━━━━━━━━━━━━━

Now Again,

$:\implies$ ∠ ACB + ∠ BAC + ∠ ABC = 180° [∵ Angle sum property: sum of ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ interior angles of a triangle is 180°.]

$:\implies$ ∠ ACB + 85° + 43° = 180°

$:\implies$ ∠ ACB + 128° = 180°

$:\implies$ ∠ ABC = 180° - 128°

$:\implies{\underline{\boxed{\bf{\pink{ \angle ABC = 52^\circ}}}}}$ ★

$\therefore\sf \underline{Hence,\; \bf{ \angle BAC = 85^\circ}\;and\; \bf{ \angle ABC = 52^\circ}.}$

━━━━━━━━━━━━━━━

Answer 2

Answer:

wise ur id...............,.............