Answer: see proof below
<u>Step-by-step explanation:</u>
Given: A + B + C = π → A = π - (B + C)
→ B = π - (A + C)
→ C = π - (A + B)
Use Sum to Product Identity: sin A - sin B = 2 cos [(A + B)/2] · sin [(A - B)/2]
Use the following Cofunction Identity: cos (π/2 - A) = sin A
<u>Proof LHS → RHS:</u>
LHS: sin A - sin B + sin C
= (sin A - sin B) + sin C




![\text{Factor:}\qquad 2\sin \bigg(\dfrac{C}{2}\bigg)\bigg[ \sin \bigg(\dfrac{A-B}{2}\bigg)+\cos \bigg(\dfrac{C}{2}\bigg)\bigg]](https://tex.z-dn.net/?f=%5Ctext%7BFactor%3A%7D%5Cqquad%202%5Csin%20%5Cbigg%28%5Cdfrac%7BC%7D%7B2%7D%5Cbigg%29%5Cbigg%5B%20%5Csin%20%5Cbigg%28%5Cdfrac%7BA-B%7D%7B2%7D%5Cbigg%29%2B%5Ccos%20%5Cbigg%28%5Cdfrac%7BC%7D%7B2%7D%5Cbigg%29%5Cbigg%5D)
![\text{Given:}\qquad 2\sin \bigg(\dfrac{C}{2}\bigg)\bigg[ \sin \bigg(\dfrac{A-B}{2}\bigg)+\cos \bigg(\dfrac{\pi -(A+B)}{2}\bigg)\bigg]\\\\\\.\qquad \qquad =2\sin \bigg(\dfrac{C}{2}\bigg)\bigg[ \sin \bigg(\dfrac{A-B}{2}\bigg)+\cos \bigg(\dfrac{\pi}{2} -\dfrac{(A+B)}{2}\bigg)\bigg]](https://tex.z-dn.net/?f=%5Ctext%7BGiven%3A%7D%5Cqquad%202%5Csin%20%5Cbigg%28%5Cdfrac%7BC%7D%7B2%7D%5Cbigg%29%5Cbigg%5B%20%5Csin%20%5Cbigg%28%5Cdfrac%7BA-B%7D%7B2%7D%5Cbigg%29%2B%5Ccos%20%5Cbigg%28%5Cdfrac%7B%5Cpi%20-%28A%2BB%29%7D%7B2%7D%5Cbigg%29%5Cbigg%5D%5C%5C%5C%5C%5C%5C.%5Cqquad%20%5Cqquad%20%3D2%5Csin%20%5Cbigg%28%5Cdfrac%7BC%7D%7B2%7D%5Cbigg%29%5Cbigg%5B%20%5Csin%20%5Cbigg%28%5Cdfrac%7BA-B%7D%7B2%7D%5Cbigg%29%2B%5Ccos%20%5Cbigg%28%5Cdfrac%7B%5Cpi%7D%7B2%7D%20-%5Cdfrac%7B%28A%2BB%29%7D%7B2%7D%5Cbigg%29%5Cbigg%5D)
![\text{Cofunction:}\qquad 2\sin \bigg(\dfrac{C}{2}\bigg)\bigg[ \sin \bigg(\dfrac{A-B}{2}\bigg)+\sin \bigg(\dfrac{A+B}{2}\bigg)\bigg]](https://tex.z-dn.net/?f=%5Ctext%7BCofunction%3A%7D%5Cqquad%202%5Csin%20%5Cbigg%28%5Cdfrac%7BC%7D%7B2%7D%5Cbigg%29%5Cbigg%5B%20%5Csin%20%5Cbigg%28%5Cdfrac%7BA-B%7D%7B2%7D%5Cbigg%29%2B%5Csin%20%5Cbigg%28%5Cdfrac%7BA%2BB%7D%7B2%7D%5Cbigg%29%5Cbigg%5D)
![\text{Sum to Product:}\qquad 2\sin \bigg(\dfrac{C}{2}\bigg)\bigg[ 2\sin \bigg(\dfrac{A}{2}\bigg)\cdot \cos \bigg(\dfrac{B}{2}\bigg)\bigg]\\\\\\.\qquad \qquad \qquad \qquad =4\sin \bigg(\dfrac{A}{2}\bigg)\cdot \cos \bigg(\dfrac{B}{2}\bigg)\cdot \sin \bigg(\dfrac{C}{2}\bigg)](https://tex.z-dn.net/?f=%5Ctext%7BSum%20to%20Product%3A%7D%5Cqquad%202%5Csin%20%5Cbigg%28%5Cdfrac%7BC%7D%7B2%7D%5Cbigg%29%5Cbigg%5B%202%5Csin%20%5Cbigg%28%5Cdfrac%7BA%7D%7B2%7D%5Cbigg%29%5Ccdot%20%5Ccos%20%5Cbigg%28%5Cdfrac%7BB%7D%7B2%7D%5Cbigg%29%5Cbigg%5D%5C%5C%5C%5C%5C%5C.%5Cqquad%20%5Cqquad%20%5Cqquad%20%5Cqquad%20%3D4%5Csin%20%5Cbigg%28%5Cdfrac%7BA%7D%7B2%7D%5Cbigg%29%5Ccdot%20%5Ccos%20%5Cbigg%28%5Cdfrac%7BB%7D%7B2%7D%5Cbigg%29%5Ccdot%20%5Csin%20%5Cbigg%28%5Cdfrac%7BC%7D%7B2%7D%5Cbigg%29)

From the diagram, all three angles are 90° (Diagonals of a Rhombus)
Angle 1 = 9x = 90
⇒ 9x = 90
x = 90/9
x = 10
Angle 2 = y + x = 90
⇒ y + x = 90
since x = 10,
y + 10 = 90
y = 90 - 10
y = 80
Angle 3 = 15 z = 90
⇒ 15z = 90
z = 90/15
z = 6
Answer is Option C
<span>C. x = 10, y = 80, z = 6</span>
It might be C. I used the process of elimination, so I'm not so certain
Answer:
The principal investment required to get a total amount of $ 1,000,000.00 from compound interest at a rate of 6% per year compounded 12 times per year over 45 years is $ 67,659.17.
Step-by-step explanation:
Given
- Accrued Amount A = $1000000
- Interest rate r = 6% = 0.06
- Compounded monthly n = 12
To determine:
Using the formula


substituting A = 1000000, r = 0.06, t = 45, and n = 12


$
Therefore, the principal investment required to get a total amount of $ 1,000,000.00 from compound interest at a rate of 6% per year compounded 12 times per year over 45 years is $ 67,659.17.
Answer:
c)
Step-by-step explanation:
16) Cost price = Selling price * 100 / (100- loss%)
= 800 * 100 / 80 = 10*100 = 1000
Loss = Cost Price - Selling price = 1000 -800 = 200