Question

In Fig. 6.31, if PQ II ST. Z PQR = 110° and ZRST= 130°, find ZQRS. [Hint : Draw a line parallel to ST through point R.)

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Answer 1

${\tt{Question}}\: \purple☟$

In Fig. 6.31, if PQ II ST. Z PQR = 110° and ZRST= 130°, find ZQRS. [Hint : Draw a line parallel to ST through point R.)

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$\red&#9998{\tt{Answer}}\: \orange☟$

⠀⠀$&#9679\purple{\bf{See \:the \:attachment}}\red{⇑}$

⠀ $&#9679\orange{\bf{Question\: solved}}\: \green✔$

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All necessary formulas⤵️

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$\purple{\boxed{\bf{Angle\:sum\: property}}}$

Angle sum property of triangle states that the sum of interior angles of a triangle is 180°. Proof .Thus, the sum of the interior angles of a triangle is 180°.

$\blue{\tt{Example:-}}$

$\red{\boxed{a+b+c=180°}}$

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$\orange{\boxed{\bf{Alternate\: interior\:angle}}}$

Alternate interior angles are angles formed when two parallel or non-parallel lines are intersected by a transversal.Alternate interior angles are equal if the lines intersected by the transversal are parallel.

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$\orange\star{\bf{\red{\underbrace{complementary \:angle}}}}\red\star$

The sum of 2 numbers$=90°$

example $a−b=90°$

how to find "a" if a is not mentioned

$\red{\underbrace{\bf{\orange{Given࿐}}}}$

$a= \: ?$

$b = 40$

$a+40=\:90°$

$a=90-40°$

$a=50°$

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$\pink\star{\bf{\purple{\underbrace{supplementary\: angle}}}}\red\star$

The sum of two numbers$= \:180°$

example $a+b=180°$

how to find "a" if a is not mentioned

Given

$a= \:?$

$b =\: 40$

$a+40=180°$

$a=180-40°$

$a=140°$

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$\orange\star{\bf{\green{\underbrace{Adjacent \:angle}}}}\red\star$

If there is a common ray between ${\bf&#x2220}a$ and ${\bf&#x2220}b$ so it is a adjacent angle.

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$\orange\star{\bf{\blue{\underbrace{Vertical\: opposite\: angle }}}}\red\star$

Vertical angles are pair angles formed when two lines intersect. Vertical angles are sometimes referred to as vertically opposite angles because the angles are opposite to each other.

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$\orange\star{\bf{\orange{\underbrace{lenear\: pair \:of\: angles}}}}\red\star$

Here ${\bf&#x2220}a$$+{\bf&#x2220}b$$=$180°.

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Answer 2

SOLUTION:

Let us draw a parallel line XY to PQ || ST and passing through point R.

Sum of interior angle on the same side of the transversal is always = 180°

SO THAT

∠ PQR + ∠ QRX = 180°

Given that ∠ PQR= 110°

110° + ∠QRX = 180°

∠QRX = 180° -110°

∠QRX = 70°

Sum of interior angle on the same side of the transversal is always = 180°

∠RST + ∠SRY = 180° (Co-interior angles on the same side of transversal SR)

Also

130° + ∠SRY = 180°

∠SRY = 50°

XY is a straight line. Use property of linear pair we get

∠QRX + ∠QRS + ∠SRY = 180°

70° + ∠QRS + 50° = 180°

∠QRS = 180° − 120°

= 60°