Given:

FIGURE in the attachment

To find:

The value of x

Solution:

Let BD and AE intersect at O.

Step 1:

In Δ AOB, we have

∠OAB + ∠OBA + ∠AOB = 180° ..... [Angle sum property]

⇒ 70° + 60° + ∠AOB = 180°

⇒ ∠AOB = 180° - 130°

⇒ ∠AOB = 50°

Step 2:

Also,

∠AOB + ∠AOD = 180° ..... (i) .... [Linear Pairs]

⇒ 50° + ∠AOD = 180°

⇒ ∠AOD = 180° - 50°

⇒ ∠AOD = 130° ...... (ii)

and

∠AOD + ∠DOE = 180° ..... (iii) ..... [Linear Pairs]

From (i) & (iii), we get

∠AOB + ∠AOD = ∠AOD + ∠DOE

⇒ ∠AOB = ∠DOE = 50° ..... (iv)

Step 3:

In Δ AOD, we have

∠DAO + ∠AOD + ∠ODA = 180° ..... [Angle sum property]

substituing from (ii) and given figure

⇒ 10° + 130° + ∠ODA = 180°

⇒ ∠ODA = 180° - 140°

⇒ ∠ODA = 40° ...... (v)

Step 4:

∠OAD + ∠AOD = ∠ODC ..... [Sum of TWO opposite interior angles of a triangle is EQUAL to its exterior angle]

⇒ 10° + 130° = ∠ODC

⇒ ∠ODC = 140° ..... (vi)

Step 5:

∴ ∠ODE = ∠ODC - ∠ODA

substituting from (v) & (vi)

⇒ ∠ODE = 140° - 40°

⇒ ∠ODE = 100° ..... (vii)

Step 6:

Now,

In Δ ODE, we have

∠ODE + ∠DOE + ∠DEO = 180°

substituting from (iv) & (vii)

⇒ 100° + 50 + x = 180°

⇒ x = 180° - 150°

⇒ x = 30°

Thus, the value of x is 30°.

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