SAT Angles

Correct Answer: D

Solution:

Tip: Angles at a point sum to $360^\circ.$
Tip: Angles on a line sum to $180^\circ.$
Using the diagram and the given, we form a system of linear equations:

$5a^\circ + 9b^\circ = 360^\circ \quad \text{Angles at a point sum to}\ 360^\circ. \qquad (1)$

$5a^\circ + 3b^\circ = 180^\circ \quad \text{Angles on a line sum to}\ 180^\circ. \qquad (2)$

Subtracting $(2)$ from $(1),$ we obtain:

$\begin{array}{r c l l} 5a^\circ + 9b^\circ - (5a^\circ + 3b^\circ) &=& 360^\circ - 180^\circ &\quad \text{subtract} (2)\ \text{from}\ (1)\\ 5a^\circ + 9b^\circ - 5a^\circ - 3b^\circ &=& 180^\circ &\quad \text{distribute}\\ 6b^\circ &=& 180^\circ &\quad \text{combine like terms}\\ b^\circ &=& 30^\circ &\quad \text{divide both sides by}\ 6\\ \end{array}$

Plugging $b=30^\circ$ into equation $(2),$ we solve for $a^\circ.$

$\begin{array}{r c l l} 5a^\circ + 3b^\circ &=& 180^\circ &\quad \text{use equation}\ (1)\\ 5a^\circ + 3\cdot 30^\circ &=& 180^\circ &\quad \text{plug in}\ b=30^\circ\\ 5a^\circ + 90^\circ &=& 180^\circ &\quad 3 \cdot 30 =90\\ 5a^\circ &=& 90^\circ &\quad \text{subtract}\ 90^\circ\ \text{from both sides}\\ a^\circ &=& 18^\circ &\quad \text{divide both sides by}\ 5\\ \end{array}$

We're looking for the sum of $a^\circ$ and $b^\circ:$

$a^\circ + b^\circ = 18^\circ + 30^\circ = 48^\circ.$

Note: We cannot assume that $a^\circ + a^\circ + a^\circ + 2a^\circ = 90^\circ$ even though, coincidentally, this happens to be the case. We can only use what is given.

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Incorrect Choices:

(A)
Tip: Read the entire question carefully.
If you solve for $b^\circ-a^\circ$ instead of $a^\circ+b^\circ,$ you will get this wrong answer.

(B)
If you add the coefficients of the unknowns that appear in the diagram, you will get this wrong answer.

(C)
Tip: Read the entire question carefully.
If you solve for $b^\circ$ instead of $a^\circ+b^\circ,$ you will get this wrong answer.

(E)
This answer is meant to prevent you from estimating. The diagram is drawn to scale, so you could eliminate choices (A), (B), and (C) because $a^\circ+b^\circ,$ seems to be greater than $30^\circ.$ But you wouldn't be able to eliminate (E).

By looking very carefully on the right side of the given figure we found that the value of a+a+a+2a=90 degrees so 5a =90 degrees then a=90÷5=18 degrees now on the same side we can also observe that b+2b=90 degrees then b =90÷3=30 degrees hence a degree + b degree = 18 +30 = 48 degrees which us is our answer

4b+2b=180 (straight line) b=30 a+a+a+2a+b+2b=180 (straight line) 5a+30+2*30=180 5a=180 - 90 a=90/5 a=18 a+b=30+18=48

Since AB is a straight line and it's perpendicular to the horizontal line.

$5a=90$ then $a=18$

$3b=90$ then $b=30$

$a+b=48$

Since $AB$ is a straight line, we see that $6b^\circ=180^\circ$ and $5a^\circ+3b^\circ=180^\circ$
So, $6b^\circ=180^\circ\ \Rightarrow b^\circ=30^\circ$
$5a^\circ+3b^\circ=180^\circ\ \Rightarrow 5a^\circ+90^\circ = 180^\circ \Rightarrow a^\circ= 18^\circ$
Therefore, $a^\circ+b^\circ = 18^\circ + 30^\circ = 48^\circ$

First, take the first quadrant. In this add the all a's. So adding these and then equating them to 90° we get, 5a=90 a=18° Now, to find b take the left side. We get, 6b=180 (as AB is a straight line) b=30°

Now, add a° + b° 18° + 30° =48° And this is the final answer.

You know that 3b is equal to 90, so

90 ÷ 3 = 30

b=30

And 5a is equal to 90, so

90 ÷ 5 = 18

a = 18

a + b = ?

18 + 30 = 48

Answer

D) 48

4b+2b=180 then 6b=180 then b=30/ a+a+a+2a=90 then 5a=90 then a=18

Since AB is a strait line, we see to the left of the line that 4b +2b = 180. Therefore 6b = 180, and b = 30. Let's move on to the other side of the line: 5a + 3b = 180, fill in the 30 at b to get 5a + 90 = 180 and 5a = 90, divide for a = 15. Solving for a + b where a = 15 and b = 30, a + b = 45.

Solution :

Step 1 of 3: Finding $b ^ \circ$

If you pay attention to the left side of the diagram:

$\boxed{4b ^ \circ + 2b ^ \circ = 180 ^ \circ}$

Which can be simplified as:

$\boxed{6b ^ \circ = 180 ^ \circ}$

We can then divide both $6b ^ \circ$ and $180 ^ \circ$ by 6 to get $b ^ \circ$ :

$\boxed{\frac {6b ^ \circ}{6} = \frac{180 ^ \circ}{6} \Rightarrow b ^ \circ = 30 ^ \circ }$

Step 2 of 3: Finding $a ^ \circ$

If you pay attention to the right side of the diagram:

$\boxed{a ^ \circ + a ^ \circ + a ^ \circ + 2a ^ \circ + b ^ \circ + 2b ^ \circ = 180 ^ \circ}$

Which can be simplified as:

$\boxed{5a ^ \circ + 3b ^ \circ = 180 ^ \circ}$

We have figured out $b ^ \circ$ and we don't need it right now, so move $3b ^ \circ$ to the right side of the equation then subtract $3b ^ \circ$ from $180 ^ \circ$ :

$\boxed{5a ^ \circ + 3(30) ^ \circ = 180 ^ \circ \Rightarrow 5a ^ \circ + 90 ^ \circ = 180 ^ \circ}$

$\boxed{\Rightarrow 5a ^ \circ = 180 ^ \circ - 90 ^ \circ \Rightarrow 5a^ \circ = 90 ^ \circ}$

We can then divide both $5a ^ \circ$ and $90 ^ \circ$ by 5 to get $a ^ \circ$ :

$\boxed{\frac {5a ^ \circ}{5} = \frac {90 ^ \circ}{5} \Rightarrow a ^ \circ = 18 ^ \circ}$

Step 3 of 3: Finding $a ^ \circ + b ^ \circ$

To find $a ^ \circ + b ^ \circ$ , simply sum them up:

$\boxed{a ^ \circ + b ^ \circ \Rightarrow 18 ^ \circ + 30 ^ \circ = 48 ^ \circ}$

Correct Answer: D $(48 ^ \circ)$

(I'm new to Brilliant, so please excuse my mistakes)