Question

PQ is a straight line segment.According to the information marked in the figure, (I) Find the magnitude of PÔS. (ii) Find magnitude of SÔQ.



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99 POINTS!! btw turn the attached picture upside down​

Answer 1

Answer:

magnitude of PÔS = 60°

magnitude of SÔQ = 120°

note: a straight line has always 180°

solve for x:

• PÔS + SÔQ = 180°

• 3x + 2x + 80° = 180°

• 5x = 180° - 80°

• 5x = 100°

• x  = 20°

m∠POS

• 3x

• 3(20°)

• 60°

m∠SOQ

• 2x + 80°

• 2(20°) + 80°

• 40° + 80°

• 120°

Answer 2

So to find angle POS and SOQ we have to find value of x first.

$\rm \angle POS + \angle SOR + \angle ROQ = 180 \degree$

Reason:-

Sum of angles on same line.

By the way it can be any number of angles but they should meet at same point. And their sum is always 180°. Remember linear pair? It is also same aspect but there will be only two angles.

So:-

$\dashrightarrow \sf \angle POS + \angle SOR + \angle ROQ = 180 \degree$

$\dashrightarrow \sf 3x + 2x+ 80= 180 \degree$

$\dashrightarrow \sf 5x+ 80= 180 \degree$

$\dashrightarrow \sf 5x= 180 - 80 \degree$

$\dashrightarrow \sf 5x= 100 \degree$

$\dashrightarrow \sf x= \dfrac{100}{5} \degree$

$\dashrightarrow \sf x= \dfrac{5 \times 20}{5} \degree$

$\dashrightarrow \sf x= \dfrac{\cancel5 \times 20}{\cancel5} \degree$

$\dashrightarrow \sf x= \dfrac{1 \times 20}{1} \degree$

$\dashrightarrow \bf x=20\degree$

Verification:-

$\dashrightarrow \sf 3x + 2x+ 80= 180 \degree$

Put value of x

$\dashrightarrow \sf( 3 \times 20) + (2 \times 20)+ 80= 180 \degree$

$\dashrightarrow \sf60 + (2 \times 20)+ 80= 180 \degree$

$\dashrightarrow \sf60 + 40+ 80= 180 \degree$

$\dashrightarrow \sf100+ 80= 180 \degree$

$\dashrightarrow \bf180= 180 \degree$

LHS = RHS

HENCE VERIFIED!

• Angle POS = 3x
• Angle POS = 3 × 20
• Angle POS = 60°

• Angle SOQ = 2x + 80
• Angle ROQ = 2 × 20 + 80
• Angle ROQ = 40 + 80
• Angle ROQ = 120°