Question

options niche waale hain uper waale nahi..... class 9th ch.6​

Answer 1

Solution :

Produce PQ to intersect SR in a point U.

Now,

PU || ST and transversal SU intersects them at U and S respectively.

$\therefore\sf \angle \: SUQ = \angle \: TSU\: \: (Alternate \: Angles) \:$

$\longrightarrow\sf \angle \: SUQ = {130}^{ \circ} \:$

$\longrightarrow\sf \angle \: QUR = {180 - 130}^{ \circ} = {50}^{ \circ}$

$\therefore \sf \angle \: SUQ + \angle \: QUR\: = {180}^{ \circ} \: \: (Linear\: Pairs)$

Since,ray PQ stands at Q on OU.

$\therefore \sf \angle \: PQR + \angle \: RQU = {180}^{ \circ}$

$\longrightarrow\sf {110}^{ \circ} + \angle \: RQU = {180}^{ \circ}$

$\longrightarrow\sf \angle \: RQU = {70}^{ \circ}$

In Triangle QRS :

Angle (QUR + RQU + QRS) = 180°

(Sum of all angles of Triangle is 180°)

→ 50° + 70° + QRS = 180°

→ 120° + QRS = 180°

→ QRS = 60°

$\therefore$ Value of angle QRS is 60°.

Answer 2

✴️ANSWER✴️

✍️Interior angles on the same side of the transversal:The pair of interior angles on the same side of the transversal are called consecutive interior angles or allied angles or co interior angles.

✍️If a transversal intersects two Parallel Lines then each pair of interior angles on the same side of the transversal is supplementary.

✍️If a transversal intersects two lines such that a pair of alternate interior angles is equal then the two lines are parallel.

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✴️SOLUTION :✴️

Given :

PQ || ST, ∠PQR = 110° and ∠RST = 130°

Construction:

A line XY parallel to PQ and ST is drawn.

∠PQR + ∠QRX = 180° (Angles on the same side of transversal.)

110° + ∠QRX = 180°

∠QRX = 180° - 110°

∠QRX = 70°

Also,∠RST + ∠SRY = 180° (Angles on the same side of transversal.)

130° + ∠SRY = 180°

∠SRY = 50°

Now,∠QRX +∠SRY + ∠QRS = 180°

70° + 50° + ∠QRS = 180°

∠QRS = 60°

Hence, ∠QRS = 60°

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