Question

3. In triangle PQR the sides PQ and PR have been produced to S and T respectively.
Bisectors of angle RQS and angle QRT meet at M. If <P = 78° find angle QMR​

Answer 1

Answer:

Step-by-step explanation:

Answer:

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,

So,

Therefore,

Hence, Proved.

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