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# 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, $$x+y+z+w= 60^{\circ}+80^{\circ}+120^{\circ}+100^{\circ} =360^{\circ}$$