Prove the theorem which states that the angle subtended by a cord is twice at the centre than at the circumference
1 answer:
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Explanation:
<u>Given</u>:
- The attached figure showing circle O, chord BC, central angle BOC and inscribed angle BAC
- angle BAC = α + β
<u>Prove</u>:
- angle BOC = 2×angle BAC
<u>Proof</u>:
∠BOA +∠BOC +∠AOC = 360° . . . . . sum of arcs of a circle is 360°
2α +∠BOA = 180°, 2β +∠AOC = 180° . . . . . sum of triangle angles is 180°
∠BOA = 180° -2α, ∠AOC = 180° -2β . . . . solve statement 2 for central angles
(180° -2α) +∠BOC +(180° -2β) = 360° . . . . . substitute into statement 1
∠BOC = 2(α +β) . . . . . add 2α+2β-360° to both sides
∠BOC = 2×∠BAC . . . . . substitute given for α+β; the desired conclusion
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Answer: The critical value of the test is 2.807.
Step-by-step explanation:
Since we have given that
n = 24
So, the degree of freedom v = n-1 = 24 - 1=23
Since, n>30, so we will use 't-test'.
Here, α = 0.01
so, According to t-table, we get that
If we take α, then we are taking about two tail test.
If we take then we are talking about one tail test.
But in both the cases, the result will be same.
Hence, the critical value of the test is 2.807.