Question

the sides PQ and PR are produced to S and T respectively.if the bisectors of angle SQR and angle TRQ intersect at O,then prove angle QOR=90-1/2 angle P.O is outside of triangle PQR.​

Answer 1

Answer:

any figure is there,

then plz provide that it would be easy

Answer 2

Answer:

$Angle\ QOR=90-\frac{1}{2}(Angle\ P)$

Explanation:

In the question,

The triangle PQR as shown in the figure,

The sides PQ and PR are extended to S and T.

QO and RO are bisectors.

To prove : ∠QOR = 90° - (1/2)(∠P)

Proof : Let us say,

∠RQO = ∠SQO = x°

and,

∠QRO = ∠TRO = y°

So, as PQS and PRT is a line.

∠PQR + 2x = 180°

So,

∠PQR = 180° - 2x ......(1)

And,

∠PRQ + 2y = 180°

So,

∠PRQ = 180° - 2y ..........(2)

In triangle PQR, as sum of the internal angles of the triangle is 180°.

So,

∠P + ∠PQR + ∠PRQ = 180°

On putting the values from the eqn. (1) and (2) we get,

∠P = 180° - (180° - 2x) - (180° - 2y)

∠P = 2x + 2y - 180° ........(3)

So,

Also,

In triangle QRO,

∠QOR = 180° - (x + y) =180° - x - y (Because, sum of the internal angles of the triangle is 180°.)

Now,

Taking half of eqn. (3) we get,

$\frac{1}{2}(P)=\frac{1}{2}(2x+2y-180)=x+y-90$

So,

$90 - \frac{1}{2}(P)=90-(x+y-90)\\=180-x-y$

Therefore,

$Angle\ QOR=90-\frac{1}{2}(Angle\ P)$

Hence, Proved.