some young people live next door. the people are nosiy...join by relative clause
Answers
Answer:
sooooooooooooooooooooooooooooooo
Explanation:
✯
ᴀɴsᴡᴇʀ
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★ 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)➪
∠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)➪
∠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)➪
∠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)➪
∠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)➪
∠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)➪
∠ACB=70°
......(vi)
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