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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
4*4=?
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at what price a 6.25%rs 100 share be quoted when the money is worth 5%
Program of "surface of revolution" in MATLAB
ABCD is a rectangle. Δ ADE and Δ ABF are two triangles such that ∠E=∠F as shown in the figure. Prove that AD x AF=AE x AB. Ans: Consider Δ ADE and Δ ABF ∠D = ∠B
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ab=8cm,bc=6cm,ca=5cm draw an incircle.
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