Question

find the values of x y z
$angle \: a \: is \: x \: and \: outer \: is \: 115 \: b \: isy \:$
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Answer 1

i) ∠x = ∠ 55° (Vertically opposite angle)

i) ∠x = ∠ 55° (Vertically opposite angle)∠x + ∠y = 180° (Adjacent angles)

i) ∠x = ∠ 55° (Vertically opposite angle)∠x + ∠y = 180° (Adjacent angles)55° + ∠y = 180° (Linear pair angles)

i) ∠x = ∠ 55° (Vertically opposite angle)∠x + ∠y = 180° (Adjacent angles)55° + ∠y = 180° (Linear pair angles)Therefore, ∠y=180° - 55° = 125°

i) ∠x = ∠ 55° (Vertically opposite angle)∠x + ∠y = 180° (Adjacent angles)55° + ∠y = 180° (Linear pair angles)Therefore, ∠y=180° - 55° = 125°∠y = ∠z

i) ∠x = ∠ 55° (Vertically opposite angle)∠x + ∠y = 180° (Adjacent angles)55° + ∠y = 180° (Linear pair angles)Therefore, ∠y=180° - 55° = 125°∠y = ∠zHence, ∠x=55° , ∠y=125° and ∠125°

i) ∠x = ∠ 55° (Vertically opposite angle)∠x + ∠y = 180° (Adjacent angles)55° + ∠y = 180° (Linear pair angles)Therefore, ∠y=180° - 55° = 125°∠y = ∠zHence, ∠x=55° , ∠y=125° and ∠125°ii) 25° + x + 40° = 180°

i) ∠x = ∠ 55° (Vertically opposite angle)∠x + ∠y = 180° (Adjacent angles)55° + ∠y = 180° (Linear pair angles)Therefore, ∠y=180° - 55° = 125°∠y = ∠zHence, ∠x=55° , ∠y=125° and ∠125°ii) 25° + x + 40° = 180°65° + x = 180°

i) ∠x = ∠ 55° (Vertically opposite angle)∠x + ∠y = 180° (Adjacent angles)55° + ∠y = 180° (Linear pair angles)Therefore, ∠y=180° - 55° = 125°∠y = ∠zHence, ∠x=55° , ∠y=125° and ∠125°ii) 25° + x + 40° = 180°65° + x = 180°Therefore, x = 180° - 65° = 115°

i) ∠x = ∠ 55° (Vertically opposite angle)∠x + ∠y = 180° (Adjacent angles)55° + ∠y = 180° (Linear pair angles)Therefore, ∠y=180° - 55° = 125°∠y = ∠zHence, ∠x=55° , ∠y=125° and ∠125°ii) 25° + x + 40° = 180°65° + x = 180°Therefore, x = 180° - 65° = 115°40° + y = 180° (Linear pairs)

i) ∠x = ∠ 55° (Vertically opposite angle)∠x + ∠y = 180° (Adjacent angles)55° + ∠y = 180° (Linear pair angles)Therefore, ∠y=180° - 55° = 125°∠y = ∠zHence, ∠x=55° , ∠y=125° and ∠125°ii) 25° + x + 40° = 180°65° + x = 180°Therefore, x = 180° - 65° = 115°40° + y = 180° (Linear pairs)Therefore y = 180° - 40° = 140°

i) ∠x = ∠ 55° (Vertically opposite angle)∠x + ∠y = 180° (Adjacent angles)55° + ∠y = 180° (Linear pair angles)Therefore, ∠y=180° - 55° = 125°∠y = ∠zHence, ∠x=55° , ∠y=125° and ∠125°ii) 25° + x + 40° = 180°65° + x = 180°Therefore, x = 180° - 65° = 115°40° + y = 180° (Linear pairs)Therefore y = 180° - 40° = 140°y + z = 180° (Linear pairs)

i) ∠x = ∠ 55° (Vertically opposite angle)∠x + ∠y = 180° (Adjacent angles)55° + ∠y = 180° (Linear pair angles)Therefore, ∠y=180° - 55° = 125°∠y = ∠zHence, ∠x=55° , ∠y=125° and ∠125°ii) 25° + x + 40° = 180°65° + x = 180°Therefore, x = 180° - 65° = 115°40° + y = 180° (Linear pairs)Therefore y = 180° - 40° = 140°y + z = 180° (Linear pairs)140° + z = 180°

i) ∠x = ∠ 55° (Vertically opposite angle)∠x + ∠y = 180° (Adjacent angles)55° + ∠y = 180° (Linear pair angles)Therefore, ∠y=180° - 55° = 125°∠y = ∠zHence, ∠x=55° , ∠y=125° and ∠125°ii) 25° + x + 40° = 180°65° + x = 180°Therefore, x = 180° - 65° = 115°40° + y = 180° (Linear pairs)Therefore y = 180° - 40° = 140°y + z = 180° (Linear pairs)140° + z = 180°Therefore z = 180° - 140° = 40°

i) ∠x = ∠ 55° (Vertically opposite angle)∠x + ∠y = 180° (Adjacent angles)55° + ∠y = 180° (Linear pair angles)Therefore, ∠y=180° - 55° = 125°∠y = ∠zHence, ∠x=55° , ∠y=125° and ∠125°ii) 25° + x + 40° = 180°65° + x = 180°Therefore, x = 180° - 65° = 115°40° + y = 180° (Linear pairs)Therefore y = 180° - 40° = 140°y + z = 180° (Linear pairs)140° + z = 180°Therefore z = 180° - 140° = 40°Hence , x - 115°

i) ∠x = ∠ 55° (Vertically opposite angle)∠x + ∠y = 180° (Adjacent angles)55° + ∠y = 180° (Linear pair angles)Therefore, ∠y=180° - 55° = 125°∠y = ∠zHence, ∠x=55° , ∠y=125° and ∠125°ii) 25° + x + 40° = 180°65° + x = 180°Therefore, x = 180° - 65° = 115°40° + y = 180° (Linear pairs)Therefore y = 180° - 40° = 140°y + z = 180° (Linear pairs)140° + z = 180°Therefore z = 180° - 140° = 40°Hence , x - 115°y = 140° and z = 40°