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Preeti Dabral 2 years, 3 months ago
Here, a = 6 - √35
⇒1/a=1/(6−√35)⇒1/a=(6+√35)/[(6+√35)(6−√35)]
[By multiplying (6 + √35) in both numerator & denominator]
⇒1/a=(6+√35)/(62−(√35)2)⇒1/a=(6+√35)(36−35)⇒1/a=(6+√35)/1⇒1/a=6+√35
Now we have to find value of a² + 1/a²
⇒a2+1/a2=(a)2+(1/a)2⇒(6−√35)2+(6+√35)2⇒(62−2×6√35+35)+(62+2×6√35+35)⇒36−12√35+35+36+12√35+35⇒72+70⇒142
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Posted by Pushkar Gajbhiye 2 years, 3 months ago
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Preeti Dabral 2 years, 3 months ago
We are given that AB∥CD,∠APQ=50∘ and ∠PRD = 127°
We need to find the value of x and y in the figure.
∠APQ = x = 50° (Alternate interior angles)
∠PRD = ∠APR = 127° (Alternate interior angles)
∠APR = ∠QPR + ∠APQ.
127° = y + 50°
⇒ y = 77°.
Therefore, we can conclude that x = 55° and y = 77°
Alternatively, 127° = y + x (because exterior angle is equal to the sum of interior opposite angles).
so, 127° = y + 50°
which gives, x = 50° and y = 77°
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Kunwar V.P Singh 2 years, 3 months ago
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