7. (a) Find x + y + z
(b) Find x + y + z + w

(a)We know that the sum of interior angles of a triangle is \(180^{\circ}\).
So, we have \(m∠1+30^{\circ}+90^{\circ}=180^{\circ}\)
\(\Rightarrow m∠1+120^{\circ}=180^{\circ}\)

\(\Rightarrow m∠1=180^{\circ}-120^{\circ}=60^{\circ}\)

So we have,\( x+90^{\circ}=180^{\circ} \quad \)[Linear pair]

\(\Rightarrow x=180^{\circ}-90^{\circ}=90^{\circ} \)

Similarly, \( y+m∠1=180^{\circ} \quad \)[Linear pair]

\(\Rightarrow y=180^{\circ}-60^{\circ}=120^{\circ} \)

And also, \( z+30^{\circ}=180^{\circ} \quad \)[Linear pair]

\(\Rightarrow z=180^{\circ}-30^{\circ}=150^{\circ} \)

So, \(x+y+z= 90^{\circ}+120^{\circ}+150^{\circ} =360^{\circ} \)

(b)We know that the sum of interior angles of a quadrilateral is \(360^{\circ\}\).

Therefore \(m∠1+120^{\circ}+80^{\circ}+60^{\circ}=360^{\circ}\)
\(\Rightarrow m∠1+260^{\circ}=360^{\circ}\)

\(\Rightarrow m∠1=360^{\circ}-260^{\circ}=100^{\circ}\)

So we have,\( x+120^{\circ}=180^{\circ} \quad \)[Linear pair]

\(\Rightarrow x=180^{\circ}-120^{\circ}=60^{\circ} \)

Similarly, \( w+m∠1=180^{\circ} \quad \)[Linear pair]

\(\Rightarrow w=180^{\circ}-100^{\circ}=80^{\circ} \)

And also, \( z+60^{\circ}=180^{\circ} \quad \)[Linear pair]

\(\Rightarrow z=180^{\circ}-60^{\circ}=120^{\circ} \)

And also, \( y+80^{\circ}=180^{\circ} \quad \)[Linear pair]

\(\Rightarrow z=180^{\circ}-80^{\circ}=100^{\circ} \)

So, \(x+y+z+w= 60^{\circ}+80^{\circ}+120^{\circ}+100^{\circ} =360^{\circ} \)