Question

Hi, how would I solve for Angle A without factoring? I don't know how to factor and always had trouble with it, but if someone can explain in-depth how to factor in this equation to find the measurement of angle A, that would be great.

Answer 1

If this figure was a parallelogram, then the opposite angles would be congruent. This would make angles B and D the same measure

angle B = angle D

x^2+20 = 7x+50

x^2+20-7x-50 = 0 .... get everything to one side

x^2-7x-30 = 0

(x-10)(x+3) = 0 ... see note below

x-10 = 0 or x+3 = 0

x = 10 or x = -3

Note: In this step is where the factoring occurs. To factor, we need to find two numbers that multiply to -30 which is the last term, and also add to -7 which is the middle coefficient. This is a trial and error process. You should find that -10 and 3 both multiply to -30 and add to -7. I suggest making a table as shown below (attached image) to list out all the possible choices.

Perhaps a much more efficient route is to use the quadratic formula.

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We found two possible solutions for x. If x = 10, then 7x+50 is 7(10)+50 = 120 which is an obtuse angle. If x = -3, then 7x+50 = 7(-3)+50 = 29 which is acute.

Assuming the diagram is drawn to scale, this means angle D is obtuse and we'll go with x = 10 and 7x+50 = 120

Angles A and D add to 180 degrees. This is true for any pair of adjacent angles in a parallelogram.

A+D = 180

A+120 = 180

A = 180-120

A = 60

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Final Answer: Angle A = 60 degrees

This answer is based on the assumption that the diagram is drawn to scale and that this quadrilateral is a parallelogram.

Answer 2

`\angle A=60\textdegree`

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`

Since we have an equation for ∠D, substitute:

`\angle A+7x+50=180`

Notice that ∠A and ∠B are also consecutive angles. So:

`\angle A+\angle B=180`

We know the equation for ∠B. Substitute:

`\angle A+x^2+20=180`

Since both equations equal 180, we can set them equal to each other:

`\angle A+7x+50=x^2+20+\angle A`

Let's subtract ∠A from both sides. This gives us:

`7x+50=x^2+20`

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`

Now, let's subtract 7x from both sides:

`0=x^2-7x-30`

Solve for x. We can factor.

Here's the trick to factoring. If we have the following:

`0=ax^2+bx+c`

The we will need to find two numbers, p and q, such that:

`p+q=b\text{ and } pq=ac`

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`

Substitute -7x for 3x-10x. This gives us:

`0=x^2+3x-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`

And factor out a -10 from the two last terms:

`0=x(x+3)-10(x+3)`

Since the expressions within the parentheses are the same, we can use grouping to acquire:

`0=(x-10)(x+3)`

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`

Solve for x:

`x=10\text{ or } x=-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`

Substitue 10 for x:

`\angle A+7(10)+50=180`

Multiply:

`\angle A+70+50=180`

Add:

`\angle A+120=180`

Subtract 120 from both sides:

`\angle A=60\textdegree`

So, in our first case, ∠A is 60°

Case II: x=-3

Again, same equation:

`\angle A+7x+50=180`

This time, substitute -3 for x. This yields:

`\angle A+7(-3)+50=180`

Multiply:

`\angle A-21+50=180`

Add:

`\angle A+29=180`

Subtract 29 from both sides:

`\angle A=151\textdegree`

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