Answer:
![\angle A=60\textdegree](https://tex.z-dn.net/?f=%5Cangle%20A%3D60%5Ctextdegree)
Step-by-step explanation:
So we have the following parallelogram and we wish to solve for ∠A.
To do so, we will need to solve for x first. Look carefully at the parallelogram...
Notice that ∠A and ∠D are consecutive angles. In other words:
![\angle A+\angle D=180](https://tex.z-dn.net/?f=%5Cangle%20A%2B%5Cangle%20D%3D180)
Since we have an equation for ∠D, substitute:
![\angle A+7x+50=180](https://tex.z-dn.net/?f=%5Cangle%20A%2B7x%2B50%3D180)
Notice that ∠A and ∠B are <em>also</em> consecutive angles. So:
![\angle A+\angle B=180](https://tex.z-dn.net/?f=%5Cangle%20A%2B%5Cangle%20B%3D180)
We know the equation for ∠B. Substitute:
![\angle A+x^2+20=180](https://tex.z-dn.net/?f=%5Cangle%20A%2Bx%5E2%2B20%3D180)
Since both equations equal 180, we can set them equal to each other:
![\angle A+7x+50=x^2+20+\angle A](https://tex.z-dn.net/?f=%5Cangle%20A%2B7x%2B50%3Dx%5E2%2B20%2B%5Cangle%20A)
Let's subtract ∠A from both sides. This gives us:
![7x+50=x^2+20](https://tex.z-dn.net/?f=7x%2B50%3Dx%5E2%2B20)
Now, we can solve for x. This is a quadratic, so let's move all the terms to one side. To start off, let's subtract 50 from both sides:
![7x=x^2-30](https://tex.z-dn.net/?f=7x%3Dx%5E2-30)
Now, let's subtract 7x from both sides:
![0=x^2-7x-30](https://tex.z-dn.net/?f=0%3Dx%5E2-7x-30)
Solve for x. We can factor.
Here's the trick to factoring. If we have the following:
![0=ax^2+bx+c](https://tex.z-dn.net/?f=0%3Dax%5E2%2Bbx%2Bc)
The we will need to find two numbers, p and q, such that:
![p+q=b\text{ and } pq=ac](https://tex.z-dn.net/?f=p%2Bq%3Db%5Ctext%7B%20and%20%7D%20pq%3Dac)
In our equation, a is 1, b is -7, and c is -30.
So, we want two numbers that sum to -7 and multiply to (1)(-30)=-30.
We can use -10 and 3. -10+3 is -7 and -10(3) is -30. So, let's substitute our b term for -10x and 3x. In other words, we have:
![0=x^2-7x-30](https://tex.z-dn.net/?f=0%3Dx%5E2-7x-30)
Substitute -7x for 3x-10x. This gives us:
![0=x^2+3x-10x-30](https://tex.z-dn.net/?f=0%3Dx%5E2%2B3x-10x-30)
This is equivalent to our old equation.
Now, we can factor. Factor out a x from the first two terms:
![0=x(x+3)-10x-30](https://tex.z-dn.net/?f=0%3Dx%28x%2B3%29-10x-30)
And factor out a -10 from the two last terms:
![0=x(x+3)-10(x+3)](https://tex.z-dn.net/?f=0%3Dx%28x%2B3%29-10%28x%2B3%29)
Since the expressions within the parentheses are the same, we can use grouping to acquire:
![0=(x-10)(x+3)](https://tex.z-dn.net/?f=0%3D%28x-10%29%28x%2B3%29)
Note that this is essentially the distribute property. If we distribute, we will get the same as above.
Zero Product Property:
![x-10=0\text{ or }x+3=0](https://tex.z-dn.net/?f=x-10%3D0%5Ctext%7B%20or%20%7Dx%2B3%3D0)
Solve for x:
![x=10\text{ or } x=-3](https://tex.z-dn.net/?f=x%3D10%5Ctext%7B%20or%20%7D%20x%3D-3)
So, we have two cases for x. Each case will yield a different answer for ∠A.
Case I: x=10
Use our original equation of:
![\angle A+7x+50=180](https://tex.z-dn.net/?f=%5Cangle%20A%2B7x%2B50%3D180)
Substitue 10 for x:
![\angle A+7(10)+50=180](https://tex.z-dn.net/?f=%5Cangle%20A%2B7%2810%29%2B50%3D180)
Multiply:
![\angle A+70+50=180](https://tex.z-dn.net/?f=%5Cangle%20A%2B70%2B50%3D180)
Add:
![\angle A+120=180](https://tex.z-dn.net/?f=%5Cangle%20A%2B120%3D180)
Subtract 120 from both sides:
![\angle A=60\textdegree](https://tex.z-dn.net/?f=%5Cangle%20A%3D60%5Ctextdegree)
So, in our first case, ∠A is 60°
Case II: x=-3
Again, same equation:
![\angle A+7x+50=180](https://tex.z-dn.net/?f=%5Cangle%20A%2B7x%2B50%3D180)
This time, substitute -3 for x. This yields:
![\angle A+7(-3)+50=180](https://tex.z-dn.net/?f=%5Cangle%20A%2B7%28-3%29%2B50%3D180)
Multiply:
![\angle A-21+50=180](https://tex.z-dn.net/?f=%5Cangle%20A-21%2B50%3D180)
Add:
![\angle A+29=180](https://tex.z-dn.net/?f=%5Cangle%20A%2B29%3D180)
Subtract 29 from both sides:
![\angle A=151\textdegree](https://tex.z-dn.net/?f=%5Cangle%20A%3D151%5Ctextdegree)
So, in our second case, ∠A is 151°
However, 151° doesn't seem likely with how the figure is drawn.
Therefore, our final answer is 60°.
And we're done!
Edit: Fixed Incorrect Answer