Ncert solutions

Ncert solutions

Grade 7

Lines and Angles | Exercise 5.1

Question 12

Q12) Find the values of the angles x, y, and z in each of the following:

i)

ii)

Solution 12:

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

∠x + ∠y = 180° (Adjacent angles)

55° + ∠y = 180° (Linear pair angles)

Therefore, ∠y=180° - 55° = 125°

∠y = ∠z

Hence, ∠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°

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subject-cta

Question 12

Q12) Find the values of the angles x, y, and z in each of the following:

i)

ii)

Solution 12:

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

∠x + ∠y = 180° (Adjacent angles)

55° + ∠y = 180° (Linear pair angles)

Therefore, ∠y=180° - 55° = 125°

∠y = ∠z

Hence, ∠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°

Still have questions? Our expert teachers can help you out

Book a free class now

Want to top your mathematics exam ?

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subject-cta
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