Question

HELP ME TO SOLVE THE ANSWER SERIOUSLY.​

Answer 1

Answer :

1)

In fig. PQ || BC

And AC,AB are transversals...

therefore,

1) Angle ACB = y

Angle QAC = Angle ACB... Alternate Angles

But Angle QAC = 80⁰

that's why Angle ACB = 80⁰

y = 80⁰

2) Angle ABC = x

Angle PAB = Angle ABC... Alternate Angles

But Angle PAB = 30⁰

that's why Angle ACB = 30⁰

x = 30⁰

3)

<PAB + <QAB = 180⁰..angle in linear pair

30⁰ + ( z + 80⁰ ) = 180⁰

30⁰ + z + 80⁰ = 180⁰

110⁰ + z = 180⁰

z = 180⁰ - 110⁰

z = 70⁰

2)

In fig PQ || BC

and AB , AC are transversals...

therefore,

1) Angle QAC = z

Angle ACB = Angle QAC.. Alternate Angles

But ACB = 90⁰

that's why Angle QAC = 90⁰

z = 90⁰

2) Angle PAB = x

Angle ABC = Angle PAB.. Alternate Angles

But Angle ABC = 60⁰

That's why Angle PAB = 60⁰

x = 60⁰

3) <A + <B + <C = 180⁰ .. sum of all < of triangle

<A + 60⁰ + 90⁰ = 180⁰

<A + 150⁰ = 180⁰

<A = 180⁰ - 150⁰

<A = 30⁰

y = 30⁰

HOPE IT HELPS YOU

Answer 2

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