Question

in the given figure PQ and RS intersect at point O, PQ amd TU intersect at M and TU and RS intersect at point N , < POR :<ROM = 1:5 and <UMO:<UMO = 2:7 then the sum of x , y and z is
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Answer 1

Answer:

answer is 400 bcoz angle POR = 30° , angle ROM=150° by linear pair of angles

angle UMQ=40° and angle UMO=140°

Now in triangle,sum of all interior angle=180°

we find angle z=110° by opposite angle property.

finally we get x,y and z

150°+140°+110°=400°

Answer 2

Answer:

x + y + z = 400°  

Step-by-step explanation:

∠POM = ∠POR + ∠ROM = 180°

∠POR : ∠ROM = 1 : 5

1a + 5a = 180°

6x = 180° ===> a = 30°

∠ROM ≅ ∠POV (vertical angles)

x = m∠POV = 5 × 30° = 150°

m∠UMO + m∠UMQ = 180° (supplementary angles)

2b + 7b = 180° ===> b = 20°

∠NMQ ≅ ∠UMO (vertical angles)

y = m∠NMQ = 7 × 20° = 140°

m∠NOM = 180° - 150° = 30°

m∠NMO = 180° - 140° = 40°

∠ONM ≅ ∠TNS

z = m∠ONM = 180° - (30° + 40°) = 110°

x + y + z = 400°