SAT Lines, Angles & Triangles Question 21 | Free SAT Prep

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Question 21·Hard·Lines, Angles, and Triangles

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In the figure, all three segments with matching tick marks have equal lengths. Which choice gives the value of $x$ ?

For a diagram with connected isosceles triangles, assign one variable to a pair of equal base angles and express every nearby angle using that variable. Use straight-line angles to move from one triangle to the next, then use the given whole angle to form one equation. Finally, check whether the question asks for an interior angle or its supplementary exterior angle.

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Identify both isosceles triangles

The matching tick marks show that triangle $ABD$ and triangle $BDC$ are both isosceles, but their equal angles are in different locations.

Use the straight line at $D$

Let $\angle BAD=\angle ADB=a$ . Because $A$ , $D$ , and $C$ are collinear, determine $\angle BDC$ in terms of $a$ .

Split the angle labeled $105^\circ$

The $105^\circ$ angle at $B$ is made of $\angle ABD$ plus $\angle DBC$ . Express both in terms of $a$ before solving.

Desmos Guide

Use angle relationships first

Algebraic angle reasoning is faster here. From the two isosceles triangles and the straight line at $D$ , derive $\angle ABC=180-\frac{3a}{2}$ , where $a=\angle BAD$ .

Find the auxiliary angle

In Desmos, graph $y=180-\frac{3x}{2}$ and $y=105$ . Click their intersection. Its first coordinate is $50$ , which represents the auxiliary angle $a$ .

Verify the exterior angle

The exterior angle is $180-\frac{a}{2}$ . In Desmos, enter $180-\frac{50}{2}$ to verify the exterior-angle measure.

Step-by-step Explanation

Use the first isosceles triangle

Let $a=\angle BAD$ . Because $AB=BD$ , triangle $ABD$ is isosceles, so $\angle ADB=a$ . Therefore, $\angle ABD=180-2a$ .

Relate the second triangle's base angles

Since $A$ , $D$ , and $C$ are on a straight line, $\angle BDC=180-a$ . Also, $BD=DC$ , so triangle $BDC$ is isosceles. Let each base angle be $b$ , so $\angle DBC=\angle BCD=b$ . Then $2b+(180-a)=180$ , giving $b=\frac{a}{2}$ .

Use the given angle at $B$

The full angle at $B$ is the sum of two adjacent angles: $\angle ABC=\angle ABD+\angle DBC$ . Thus $105=(180-2a)+\frac{a}{2}=180-\frac{3a}{2}$ . Solving gives $a=50$ , so $b=25$ .

Find the exterior angle

The interior angle at $C$ is $\angle BCD=b=25$ . Angle $x$ forms a linear pair with this angle, so $x=180-25=155$ .

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Desmos Guide

Step-by-step Explanation

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Desmos Guide

Step-by-step Explanation