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Prove that the intercept of a tangent between two parallel tangents to a circle subtends a right angle at the centre.
Since Δ ADF ≅ Δ DFC
∠ADF = ∠CDF
∴ ∠ADC = 2 ∠CDF
Similarly we can prove ∠CEB = 2∠CEF
Since || m
∠ADC + ∠CEB = 180o
⇒2∠CDF + 2∠CEF = 180o
⇒ ∠CDF + ∠CEF = 90o
In Δ DFE
∠DFE = 90o
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Can you help me with what goes into 54
Using the same mean and standard deviation as mean m = 20.1 and a standard deviation s = 5.8. Joe was informed that he scored at the 68 th percentile on the ACT, what was Joe's ap
how to sell a product
3 3/4+(1 1/49*7/10)
how do i compute an algebra number
What is the square root of -i and argument of -i Ans) argument of -i is 270 ad 1 is the square root of -i
There actually isn't a whole lot to do throughout this case. We'll find two solutions which will form a basic set of solutions and therefore our general solution will be as,
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