In the figure, ACE, BCF and DCH are straight lines and AB || HC.
Find the angles p,q,r and s.
p=45∘,q=50∘,r=65∘,s=50∘
In ΔABC, r=180∘−(50∘+65∘)=65∘ [angle sum property of a triangle]
HC||AB⇒q=50∘ [Alternative angles].
s = q [Vertically opposite angles]
Hence, s=50∘
Since BCF is a straight line,
(p+20∘)+q+r=180∘⇒p=45∘
∴p=45∘,q=50∘,r=65∘ and s=50∘ are the required values.