In equilateral triangle , point lies on segment such that Segment is drawn. Which choice gives the measure of angle ?

Solution available with step-by-step explanation

SAT Lines, Angles & Triangles Question 45 | Free SAT Prep | Aniko

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Question 45·Hard·Lines, Angles, and Triangles

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In equilateral triangle $ABC$ , point $D$ lies on segment $BC$ such that

\frac{BD}{BC}=2-\sqrt{3}.

Segment $AD$ is drawn. Which choice gives the measure of angle $DAB$ ?

When a point divides a side of an equilateral (or isosceles) triangle by an unusual ratio, a coordinate setup is often the fastest path: put the base on the $x$ -axis, use the known height of an equilateral triangle to locate the top vertex, convert the ratio into an exact coordinate for the point, and then use slopes to compute the angle between two segments. This avoids messy angle chasing and keeps the work algebraic and controlled.

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Use a coordinate setup

Because $ABC$ is equilateral, you can place $B$ and $C$ on the $x$ -axis and write coordinates for $A$ using the $30$ - $60$ - $90$ triangle height.

Translate the ratio into a coordinate

If you choose $BC=1$ , then the condition $\frac{BD}{BC}=2-\sqrt3$ tells you the $x$ -coordinate of $D$ directly.

Compare the directions of $AB$ and $AD$

Find the slopes of $AB$ and $AD$ , then use $\tan\theta=\left|\frac{m_2-m_1}{1+m_1m_2}\right|$ to get the angle between them.

Identify the angle from its tangent

After simplifying, try rewriting the tangent value by rationalizing; then compare it to $\tan(45^\circ-30^\circ)$ .

Desmos Guide

Define the points with convenient coordinates

Enter

B=(0,0)
C=(1,0)
A=(1/2, sqrt(3)/2)
D=(2-sqrt(3), 0)

Compute the slopes

Define

mAB=(y(A)-y(B))/(x(A)-x(B))
mAD=(y(A)-y(D))/(x(A)-x(D))

Compute the angle between the lines in degrees

Define

theta=abs(arctan((mAD-mAB)/(1+mAD*mAB)))*180/pi

Then look at the displayed value of theta and choose the answer option that matches it.

Step-by-step Explanation

Place the triangle on a coordinate plane

Scale does not matter, so let $B=(0,0)$ and $C=(1,0)$ . Since $ABC$ is equilateral, the third vertex is

A=\left(\frac12,\frac{\sqrt{3}}2\right).

Because $\frac{BD}{BC}=2-\sqrt{3}$ and $BC=1$ , point $D$ is

D=(2-\sqrt{3},0).

Find the slopes of $AB$ and $AD$

Slope of $AB$ :

m_{AB}=\frac{0-\frac{\sqrt3}{2}}{0-\frac12}=\sqrt3.

Slope of $AD$ :

\begin{aligned} m_{AD} &=\frac{0-\frac{\sqrt3}{2}}{(2-\sqrt3)-\frac12} \\ &=\frac{-\frac{\sqrt3}{2}}{\frac32-\sqrt3} =\frac{\frac{\sqrt3}{2}}{\sqrt3-\frac32} =\frac{\sqrt3}{2\sqrt3-3}. \end{aligned}

Rationalize the denominator:

\begin{aligned} \frac{\sqrt3}{2\sqrt3-3} &=\frac{\sqrt3(2\sqrt3+3)}{(2\sqrt3-3)(2\sqrt3+3)} \\ &=\frac{\sqrt3(2\sqrt3+3)}{12-9} =\frac{6+3\sqrt3}{3} =2+\sqrt3. \end{aligned}

So $m_{AD}=2+\sqrt3$ .

Use the angle-between-lines formula

If two non-vertical lines have slopes $m_1$ and $m_2$ , then the tangent of the angle $\theta$ between them is

\tan\theta=\left|\frac{m_2-m_1}{1+m_1m_2}\right|.

Here,

\begin{aligned} \tan\theta &=\left|\frac{(2+\sqrt3)-\sqrt3}{1+\sqrt3(2+\sqrt3)}\right| \\ &=\frac{2}{1+2\sqrt3+3} =\frac{2}{4+2\sqrt3} =\frac{1}{2+\sqrt3}. \end{aligned}

Rationalize:

\frac{1}{2+\sqrt3}=\frac{2-\sqrt3}{(2+\sqrt3)(2-\sqrt3)}=2-\sqrt3.

Match the tangent value to a standard angle

Using $15^\circ=45^\circ-30^\circ$ ,

\tan 15^\circ=\tan(45^\circ-30^\circ)=\frac{1-\frac{1}{\sqrt3}}{1+\frac{1}{\sqrt3}}=\frac{\sqrt3-1}{\sqrt3+1}=2-\sqrt3.

So $\tan\theta=2-\sqrt3$ means $\theta=15^\circ$ . Therefore, the measure of angle $DAB$ is $15^\circ$ .

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Step-by-step Explanation

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Step-by-step Explanation