Prove that step by step

Mathematics

1 answer:

Elza [17]2 years ago

3 0

Answer:

In ΔECB ,

∠ECD = ∠EBC + ∠BEC (∵ Exterior Angle Property of a triangle)

⇒ ∠BEC = ∠ECD - ∠EBC ................. eqn.1

In ΔABC ,

∠ACD = ∠BAC + ∠ABC (∵ Exterior Angle Property of a triangle )

⇒ 2∠ECD = ∠BAC + 2∠EBC

⇒ ∠BAC = 2∠ECD - 2∠EBC = 2(∠ECD - ∠EBC)

( Putting eqn.1 here )

⇒ ∠BAC = 2∠BEC (∵∠BEC = ∠ECD - ∠EBC)

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If f is a scalar field and F, G are vector fields, then f F, F · G, and F × G are defined by the following. (f F)(x, y, z) = f(x

valentina_108 [34]

Answer:

Step-by-step explanation:

Consider curl $(fF)$ where $f$ is a scalar function and F is a vector function

$F=F_{1}i +F_{2}j +F_{3}k$

i j k

$curl(fF) = \frac{\partial}{\partial x}$ $\frac{\partial}{\partial y}$ $\frac{\partial}{\partial z}$

$fF_{1}$ $fF_{2}$ $fF_{3}$

$curl(fF)=i(\frac{\partial}{\partial y}(fF_{3})-\frac{\partial}{\partial z}(fF_{2}))-j(\frac{\partial}{\partial x}(fF_{3})-\frac{\partial}{\partial z}(fF_{1}))+k(\frac{\partial}{\partial x}(fF_{2})-\frac{\partial}{\partial y}(fF_{1}))$

$=i(\frac{\partial}{\partial y}(F_{3})+\frac{\partial}{\partial y}(f)-\frac{\partial}{\partial z}(F_{2})-\frac{\partial}{\partial z}(f))-j(\frac{\partial}{\partial x}(F_{3})+\frac{\partial}{\partial x}(f)-\frac{\partial}{\partial z}(F_{1})-\frac{\partial}{\partial z}(f))+k(\frac{\partial}{\partial x}(F_{2})+\frac{\partial}{\partial x}(f)-\frac{\partial}{\partial y}(F_{1})-\frac{\partial}{\partial y}(f))$