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Answer :
We have, \(∠y = 40^{\circ}\) (Alternate angles)
\(∠z + 40°^{\circ} = 70^{\circ}\) (Exterior angle property)
\(\Rightarrow ∠z = 70^{\circ} – 40^{\circ} = 30^{\circ}\)
\(z = ∠EPH \quad\)[Alternate angle]
In ∆EPH,
\(∠x + 40^{\circ} + ∠z = 180^{\circ}\quad\) [Adjacent angles]
\(\Rightarrow ∠x + 40^{\circ} + 30^{\circ} = 180^{\circ}\)
\(\Rightarrow ∠x + 70^{\circ} = 180^{\circ}\)
\(\Rightarrow ∠x = 180^{\circ} – 70^{\circ} = 110^{\circ}\)
Hence we have,\( x = 110^{\circ}, y = 40^{\circ} and\; z = 30^{\circ}.\)