The electric field is the negative of the gradient of the electric potential:
At the point (1.0, 1.0, 1.0), the electric field has magnitude
Compound Interest
1 answer:
You might be interested in
Answer:
∠1 = 43°, ∠2 = 17°, ∠3 = 120°, ∠4 = ∠5 = ∠6 = 60°, ∠7 = 17°, ∠8 = 120°, ∠9 = 43°.
Step-by-step explanation:
I found the figure for this problem.
It is provided that ΔABC is isosceles and ΔDBE is equilateral.
Isosceles Triangle:
According to the definition the length of two sides are equal and hence the two base angles are equal.
Equilateral Triangle:
All the sides are of equal length and can also be defined as equiangular with all the angles of measure 60°
CALCULATION:
For any triangle the sum of interior angles is always 180°.
ΔABC: ∠1+ (∠2 + ∠5 + ∠7)+∠9= 180°,
where, (∠2 + ∠5 + ∠7) is the total angle ∠ABC
ΔABC is isosceles so,
∠1 = ∠9,
4x+3 = 9x-47,
50 = 5x ⇒ x=10,
Putting the value of x in ∠1 we get,
∠1 = 4*(10)+3 = 43°,
∠1 = 43°.
ΔABD: ∠1+ ∠2+∠3= 180°
Now in order to find ∠2, firstly we have to find ∠3 in ΔABD.
According to the exterior angle property, the exterior angle of an equilateral triangle is always 120°. Since ∠3 is an exterior angle for the equilateral triangle ΔDBE therefore,
∠3 = 120°,
furthermore,
∠2 = 180° - ∠3 - ∠1,
⇒ ∠2 = 180° - 120° - 43°
∠2 = 17°.
ΔDBE: ∠4+ ∠5+∠6= 180°,
Also, ΔDBE is equilateral ⇒ ∠4 = ∠5 = ∠6 = 60°.
Now from ∠1+ (∠2 + ∠5 + ∠7)+∠9= 180°
we can find ∠7,
⇒ ∠1+ (∠2 + ∠5 + ∠7)+∠9= 180°
∠7 = 180° - (∠2 + ∠5 + ∠1+∠9),
∠7 = 180° - (17° + 60° + 43° + 43°),
∠7 = 180° - (163°)
∠7 = 17°.
Similarly, ∠8 is an exterior angle for the equilateral triangle ΔDBE therefore,
∠8 = 120°,
