Given: PK and PE tangents

m∠KPE = 60°,

2KP = PE +1

Find: EK

Step-by-step explanation:

Here we have KP = PE (tangents from outside the circle)

Therefore, 2KP = PE + 1 = KP + 1

Hence, 2KP - KP = 1 or KP = 1 = PE

Since KE is the base of triangle KPE, where  ∡ KPE = 60, and KP = PE, we have an isosceles triangle such that ∡PKE = ∡PEK

Hence, in ΔKPE, ∡KPE + ∡PKE + ∡PEK  = 180

Therefore, 60° + ∡PKE + ∡PEK  = 180

Hence, ∡PKE + ∡PEK  = 180° - 60° = 120°

Because ∡PKE = ∡PEK, (base angles of isosceles triangle), we have;

∡PKE + ∡PEK = 2·∡PEK = 120° which gives

∡PEK = 60° = ∡PKE

Therefore, ∡KPE = ∡PEK = ∡PKE = 60°

Hence, ΔKPE is an equilateral triangle and KP = PE = EK = 1

EK = 1.

  n=   0.0000 - 0.3750 i  

  n=   0.0000 + 0.3750 i

step-by-step explanation:

one of these should be correct

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step-by-step explanation: