Question

solve this problem..

Answer 1

solution :

1)
DE ll AB ....given
LDEA =LEBA.... alternate angles
LDEA=55
LEBA= 55
2)
LDEC+LECD+LCDE=180..triangle property
55+LECD+40=180
55+ LECD=180-40
55+ LECD=140
LECD=140-55
LECD=85
3)
sum of measures of all angles of triangle is 180
LABC+LBCA+LCAB= 180
LABC+85+55=180
LABC+140 =180
LABC=180-140
LABC=40

Answer 2

Q10

Given

DE║AB

Solution

DE║AB and DB is the transversal

∠EDC = ∠a [Alternate Interior angles]

∠a = 40°

DE║AB and EA is the transversal

∠DEC = ∠b [Alternate Interior Angles]

∠b = 50°

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Q11

Given

∠1 = 65°

∠2 = 65°

Solution

For two lines to be parallel, [acording to the given figure], few criterias must be proved.

→ Alternate Angles must be equal.

→ Co- Interior Angles must be supplementary.

→ Corresponding angles must be equal.

First, Lets try proving if alternate angles 1 and 3 are equal.

∠3 + ∠2 = 180° [Linear Pair]

∠3 = 180° - 65°

∠3 = 115°

Since alternate angles are not equal, we cant say that 'p' is parallel to 'q'