SAT Special Right Triangles

Correct Answer: E

Solution:

Tip: The two acute angles in a right triangle are complementary.
Tip: Know the $30^\circ-60^\circ-90^\circ$ and the $45^\circ-45^\circ-90^\circ$ Triangles.
Because $AB = AD + DB,$ we must find the lengths of $\overline{AD}$ and $\overline{DB}.$

The figure indicates that $\triangle ADC$ is a right triangle. The acute angles in a right triangle add to $90^\circ$ and since $m\angle CAD = 30^\circ, m\angle ACD = 60^\circ.$

In a $30^\circ - 60^\circ - 90^\circ$ triangle the longer leg is $\sqrt{3}$ times as long as the shorter leg. Therefore,

$AD = \sqrt{3}\cdot CD = a\sqrt{3}.$

$\angle ADC$ and $\angle CDB$ are supplementary. Therefore, their measures add to $180^\circ,$ and because $\angle ADC = 90^\circ, \angle CDB = 90^\circ$ also. Again we use the fact that the acute angles in a right triangle add to $90^\circ.$ Since $m\angle DBC = 45^\circ, m\angle DCB = 45^\circ.$

Because $\triangle CDB$ is an isosceles triangle, it follows that

$DB = CD = a.$

We can now find the length of $\overline{AB}.$

$AB = AD + DB = a\sqrt{3} + a = a+ a\sqrt{3}.$

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Incorrect Choices:

(A)
Tip: Read the entire question carefully.
This is the length of $\overline{AD},$ not of $\overline{AB}.$

(B)
Tip: Read the entire question carefully.
If you find $CD + DB$ instead of $AB$ or if you think that $AD = DB,$ you will get this wrong answer.

(C)
Tip: Know the $30^\circ-60^\circ-90^\circ$ and the $45^\circ-45^\circ-90^\circ$ Triangles.
In a $30^\circ - 60^\circ - 90^\circ$ triangle the longer leg is $\sqrt{3}$ times as long as the shorter leg. Since in $\triangle ADC,$ the shorter leg is $a,$ the longer leg $AD$ will be $a\sqrt{3},$ not $\sqrt{3}.$

(D)
Tip: Know the $30^\circ-60^\circ-90^\circ$ and the $45^\circ-45^\circ-90^\circ$ Triangles.
If you mix up the $30^\circ - 60^\circ - 90^\circ$ and $45^\circ - 45^\circ-90^\circ$ triangles, and you think that the longer leg of the $30^\circ - 60^\circ - 90^\circ$ triangle is $\sqrt{2}$ times longer than the shorter leg (instead of $\sqrt{3}$ times longer than the shorter leg), you will get this wrong answer.