Question

Find:

angle EDC
angle ADE
angle AED
angle DEC
angle DCE
angle ACB​

Answer 1

$\huge ✯{ \underline{ \underline{\bold{ᴀɴsᴡᴇʀ}}}}✯$

★ given →

• ∠BAC = 55°
• ∠BDC = 100°
• ∠BCD = 25°

and we know that Arrows on lines are used to indicate that those lines are parallel.

• ➪ DE || BC

★ solution →

i) ∠EDC

☞︎︎︎ we have DE || BC

here, DC is transversal line so ∠BCD and ∠EDC are alternate angles.

➪ ∠BCD = ∠EDC (alternate angles are equal)

and it is give that ∠BCD = 25°

$➪ \: \boxed{ \pink{\bold{ ∠EDC = 25°}}}......(i)$

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ii) ∠ADE

☞︎︎︎ we have, ACB is a line and ∠BDC, ∠EDC and ∠ADE are angles of a line.

➪ ∠BDC + ∠EDC + ∠ADE = 180°

(by linear pair or we know that sum of angles of a line is 180°)

and we have,

• ∠BDC = 100° (given)
• ∠EDC = 25° [from eq.(i)]

➪ 100° + 25° + ∠ADE = 180°

➪ 125° + ∠ADE = 180°

➪ ∠ADE = 180° - 125°

$➪ \: \boxed{ \pink{\bold{ ∠ADE = 55°}}}......(ii)$

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iii) ∠AED

☞︎︎︎ in ∆ADE

∠A + ∠D + ∠E = 180° (angle sum property of a triangle, sum of all angles of a triangle is 180°)

➪ ∠DAE + ∠ADE + ∠AED = 180°

we have,

• ∠DAE = 55° (given)
• ∠ADE = 55° [from eq.(ii)]

➪ 55° + 55° + ∠AED = 180°

➪ 110° +∠AED = 180°

➪ ∠AED = 180° - 110°

$➪ \: \boxed{ \pink{ \bold{ ∠AED = 70°}}}......(iii)$

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iv) ∠DCE

☞︎︎︎ in ∆ADC,

∠A + ∠D +∠C = 180° (angle sum property of a triangle, sum of all angles of a triangle is 180°)

➪ ∠DAC + ∠ADC + ∠DCA = 180°

➪ ∠DAC + ∠ADE + ∠EDC + ∠DCA = 180°

we have,

• ∠DAC = 55° (given)
• ∠ADE = 55° [from eq.(ii)]
• ∠EDC = 25° [from eq.(i)]

➪ 55° + 55° + 25° + ∠DCA = 180°

➪ 110° + 25° + ∠DCA = 180°

➪ 135° + ∠DCA = 180°

➪ ∠DCA = 180° - 135°

➪ ∠DCA = 45°

$➪ \: \boxed{ \pink{ \bold{ ∠DCE = 45°}}}......(iv)$

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v) ∠DEC

☞︎︎︎ in ∆DEC,

∠D + ∠E + ∠C = 180° (angle sum property of a triangle, sum of all angles of a triangle is 180°)

➪ ∠EDC + ∠DCE + ∠DEC = 180°

we have,

• ∠EDC = 25° [from eq.(i)]
• ∠DCE = 45° [from eq.(iv)]

➪ 25° + 45° + ∠DEC = 180°

➪ 70° + ∠DEC = 180°

➪ ∠DEC = 180° - 70°

$➪ \: \boxed{ \pink{ \bold{ ∠DEC = 110°}}}......(v)$

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vi) ∠ACB

☞︎︎︎ ∠ACB = ∠ACD + ∠BCD

we have,

• ∠ACD = ∠DCE = 45° [from eq.(iv)]
• ∠BCD = 25° (given)

➪ ∠ACB = 45° + 25°

$➪ \: \boxed{ \pink{\bold{ ∠ACB = 70°}}}......(vi)$

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