Question

Please help me to solve this question . the one who will give the right answer I will mark him brainliest. But please dont prank with me

Answer 1

Gɪᴠᴇɴ :-

• PR > PQ.
• PS is Angle Bisector .

Tᴏ SHOW :-

• x > y ?

Fᴏʀᴍᴜʟᴀ ᴜsᴇᴅ :-

• If one side of a triangle is longer than another side, then the angle opposite the longer side will be larger than the angle opposite the shorter side.
• An exterior angle of a triangle is equal to the sum of the opposite interior angles.

Sᴏʟᴜᴛɪᴏɴ :-

Given That, PS is Angle Bisector .

So,

→ ∠QPS = ∠RPS = Let z .

And,

Given That, PR > PQ.

So,

→ ∠PQS > ∠PRS . -------- Equation (1).

__________________

Now, In ∆PSQ, we have :-

→ ∠PSQ = ∠PRS + ∠RPS ( Exterior angle = sum of Opp. interior Angles).

→ y = ∠PRS + z ---------- Equation (2).

__________________

Similarly, in ∆PSR , we have :-

→ ∠PSR = ∠PQS + ∠QPS ( Exterior angle = sum of Opp. interior Angles).

→ x = ∠PQS + z ---------- Equation (3).

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Now, From Equation (1) ,

→ ∠PQS > ∠PRS

Adding z on Both sides we get,

→ ∠PQS + z > ∠PRS + z

Therefore, From Equation (1) & (2) , we get,

→ x > y . (Hence, Proved).

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Answer 2

GIVEN:

• PR > PQ
• PS bisects QR

TO PROVE:

• x > y = Angle PSR > Angle PSQ

SOLUTION:

PS is angular bisector so, PR > PQ &

Angle PQS > angle PRS

Assuming as equation 1

_________________

Now, In ∆PSQ, we have :

Using exter angle = sum of two interior opposite angles

→ Angle PSQ = Angle PRS + Angle RPS

→ y = Angle PRS + z

Assuming as Equation (2)

__________________________

Similarly, in ∆PSR, we have :

→ Angle PSR = Angle APS + Angle APS ( Exterior angle = sum of two opposite interior Angles)

→ x = angle PQS+z

Assuming as Equation (3).

__________________________

Now, From Equation (1),

→ Angle PQS > angle PRS

Adding z on Both sides we get,

→ angle PQS + z > angle PRS + z

From equation 1 & 2

We have

x > y

Hence, proved