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16. Let y be a circle with center O and let P, Q, and R be three points on y. Prove that if P and R are diametrically opposite, then \(\angle PQR\) is a right angle, and if O and Q are on the same side of PR, then $\(\angle PQR\)^\circ = \frac{1}{2} (\angle POR)^\circ$. (Hint: Again use the fact that the triangu- lar angle sum is 180°. There are four cases to consider, as in Fig- ure 5.18.) State and prove the analogous result when O and Q are on opposite sides of PR.

          16. Let y be a circle with center O and let P, Q, and R be three points
on y. Prove that if P and R are diametrically opposite, then \(\angle PQR\)
is a right angle, and if O and Q are on the same side of PR, then
$\(\angle PQR\)^\circ = \frac{1}{2} (\angle POR)^\circ$. (Hint: Again use the fact that the triangu-
lar angle sum is 180°. There are four cases to consider, as in Fig-
ure 5.18.) State and prove the analogous result when O and Q are
on opposite sides of PR.
        
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16. Let y be a circle with center O and let P, Q, and R be three points
on y. Prove that if P and R are diametrically opposite, then ∠ PQR
is a right angle, and if O and Q are on the same side of PR, then
∠ PQR^∘ = (1)/(2) (∠ POR)^∘. (Hint: Again use the fact that the triangu-
lar angle sum is 180°. There are four cases to consider, as in Fig-
ure 5.18.) State and prove the analogous result when O and Q are
on opposite sides of PR.

Added by Christopher C.

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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Euclidean geometry, which means you are allowed to use the parallel postulate and its consequences already established 16. Let y be a circle with center O and let P, Q, and R be three points on y. Prove that if P and R are diametrically opposite, then PQR is a right angle, and if O and Q are on the same side of PR, then (PQR) = 1/(XPOR). (Hint: Again use the fact that the triangu- lar angle sum is 180. There are four cases to consider, as in Fig ure 5.18.) State and prove the analogous result when O and Q are on opposite sides of PR. P R Q R 0 O P R 0 0 0 R
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Transcript

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00:01 Let it p m bar is equal to mb bar so p m bar divided by m b is equal to alpha divided by beta so the point m divided the line p b b p q bar in the ratio point m divides the line pq bar in the ratio alpha is to beta so given position vector of day and are p vector and q vector respectively with reference to 0 so op vector is equal to p vector and o q vector is equals to q vector so now and given point of m b m vector so o m vector would be equal to m vector now…
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