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Find angleP in the figure below....

Find `angleP` in the figure below.

Text Solution

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In ` triangleABC and triangleQRP` ,we have
`(AB)/(QR) = (3.6)/(7.2) = 1/2 , (BC)/(RP) = 6/12 = 1/2`
and ` (CA)/(PQ)= (3sqrt3)/(6sqrt3)=1/2`
Thus `(AB)/(QR)= (BC)/(RP)= (CA)/(PQ)`
Hence , by SSS criterion .
`triangleABC ~ triangleQRP (("first two")/("first two")= ("last two")/("last two")= ("first and last")/("first and last")) `
`angleC= angleP ` ( thrid angle = third angle)
But ` angleC= 180- (angleA + angleB)= 180^(@) - ( 70^(@)+60^(@)) = 50^(@)`
` angleP = 50^(@)`
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