Question

In the given figure, line l ll line m and PQ is a transversal. If √PEB =70°, find the measure of √PEA , √AEF , √BEF , √EFC , √EFD , √DFQ and √CFQ.​

Answer 1

Step-by-step explanation:

Since PQ is a transversal,

PEA = 180 - 70 = 110

AEF = PEB = 70

BEF = AEP = 110

EFC = BEF = 110

EFD = 180 - 110 = 70

DFQ = EFC = 110

CFQ = EFD =70

Answer 2

Answer:

Let us mark the points R and S on line n, T and U on line p, A and B on line l and C and D on line m.

Suppose the lines n and p intersect the line l at K and L respectively and line m at M and N respectively.

Since, the line l and line p intersect at L, then  

∠KLN=∠BLT (Vertically opposite angles)

⇒∠KLN=45∘

Since, line l ∥ line m and line p is a transversal intersecting them at L and N, then

∠KLN+∠MNL=180∘ (Pair of interior angles on the same side of transversal is supplementary)

⇒45∘+∠a=180∘

⇒∠a=180∘−45∘=135∘

Since, the line m and line p intersect at N, then

∠DNU=∠MNL (Vertically opposite angles)

⇒∠b=∠a=135∘

Since, line n ∥ line p and line m is a transversal intersecting them at M and N, then

∠NMS=∠DNU (Corresponding angles)

⇒∠c=∠b=45∘

Step-by-step explanation: