Question

See my previous question and answer mujhe answer ni chayie phle hi kaha ha i know which class book this question belong to??
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After this report this and my earlier question plz​

Answer 1

Working out:

The above question is based on lines and angles. We should know certain terms and conditions which we need during solving the above questions. I will mention them accordingly.

GiveN:

• AB || CD
• ∠APQ = 50°
• ∠PRD = 120°

We have to find angles marked as x and y.

So, let's start solving....

❒ AB || CD, PQ is a transversal

• ∠APQ = ∠PQR (Alternate angles)

⇛ ∠APQ = ∠PQR

⇛ 50° = ∠PQR (Given)

And, ∠PQR is marked as x

⇛ x = 50°

Now In ∆PQR,

• ∠PRD is the Exterior angle

We know that, exterior angle of a triangle is equal to teh sum of interior opposite angles. So,

⇛ ∠PQR + ∠QPR = ∠PRD

⇛ x + y = ∠PRD

Plugging the values of x and ∠PRD,

⇛ 50° + y = 120°

⇛ y = 120° - 50°

⇛ y = 70°

Alternative:

Like in 1)

• ∠APR = ∠PRD (Alternate angles)

⇛ ∠APQ + ∠QPR = ∠PRD

⇛ ∠APQ + y = ∠PRD

Plugging the values of ∠APQ and ∠PRD

⇛ 50° + y = 120°

⇛ y = 70°

So, the required values of x and y respectively are:

$\huge{ \boxed{ \sf{ \red{50 \degree \: and \: 70 \degree}}}}$

And we are done !!

Answer 2

$\mathtt{\huge{\underline{\red{Question\:?}}}}$

★ In the adjoining figure , AB || CD, ∠APQ = 50° & ∠PRD = 120° , then find x & y.

$\mathtt{\huge{\underline{\green{Answer:-}}}}$

✒ The value of x = 50° & y = 70° .

$\mathtt{\huge{\underline{\orange{Explaination:-}}}}$

Given :-

• AB || CD

• ∠APQ = 50°

• ∠PRD = 120°

To Find :-

• The angles x and y.

Solution :-

According to the question,

░ AB || CD , where PQ is a transversal.

∠APQ = ∠PQR (Alternate interior angles are equal)

➝ 50° = ∠PQR (Given)

Let, ∠PQR be x° ,

➝ x = 50°

Now looking, In ∆PQR,

➝ ∠PQR + ∠QPR = ∠PRD

{The Exterior angle of a triangle is equal to teh sum of interior opposite angles}.

Let ∠QPR be y;

➝ x + y = ∠PRD..........(1)

We have ,

• x = 50°

• ∠PRD = 120°

Putting in 1 equation ,

➝ 50° + y = 120°

➝ y = 120° - 50°

➝ y = 70°

░ AB || CD , where PR is a transversal,

∠APR = ∠PRD (Alternate interior angles are equal )

➝ ∠APQ + ∠QPR = ∠PRD

➝ x + y = 120°

➝ 50° + y = 120°

➝ y = 70°

Therefore , our value of x = 50° & y = 70° .

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