few_shots: Example 1:
Problem/activity:
Find the general solution for:
$${\mathrm{cos}\left({\pi\over2}+x\right)=\mathrm{tan}{\pi\over4}}$$


Standard options: 
A. Prove the Pythagorean identity sin^(2)(θ) + cos^(2)(θ) = 1 and use it to find sin(θ), cos(θ),or tan(θ) given sin(θ), cos(θ), or tan(θ) and the quadrant of the angle.
B. Prove the addition and subtraction formulas for sine, cosine, and tangent and use them to solve problems.


Your response, in json format:
{
  "explanation": "An initial step when solving for x in this problem requires the use of the addition formula for cosine on the left side of the equation. Solving this problem does not require use of the Pythagorean identity.",
  "answer": "B"
}

Example 2:
Problem/activity:
Task
Solve the quadratic equation
$$ x^2 = (2x - 9)^2 $$
using as many different methods as possible.


Standard options: 
A. Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters.
B. Solve quadratic equations in one variable.


Your response, in json format:
{
  "explanation": "This task shows one quadratic equation with one variable, x.",
  "answer": "B"
}

Example 3:
Problem/activity:
Task
Suppose $A$ is an angle with vertex $P$, as pictured below:
[IMG]
Draw a circle with center $P$ and with  radius $r \gt 0$.  Explain why the circle
meets each ray of angle $A$ in a single point.  Label these points $Q$ and $R$ respectively.
Draw circles with centers $Q$ and $R$ respectively and radius $r$.  These circles meet at $P$ and a second point to be labelled $B$.
Show that ray $\overrightarrow{PB}$  bisects angle $P$.


Standard options: 
A. Make formal geometric constructions with a variety of tools and methods (compass and straightedge, string, reflective devices, paper folding, dynamic geometric software, etc.). Copying a segment; copying an angle; bisecting a segment; bisecting an angle; constructing perpendicular lines, including the perpendicular bisector of a line segment; and constructing a line parallel to a given line through a point not on the line.
B. Construct an equilateral triangle, a square, and a regular hexagon inscribed in a circle.


Your response, in json format:
{
  "explanation": "This task asks students to draw a ray that bisects an angle. The act of drawing circles and rays may involve compass and straightedge or dynamic geometric software.",
  "answer": "A"
}

Now, assign the following problem or activity to one or more relevant standards it aligns with.
shot_info: fl_problem_001894

HS_0

im_task_000870

HS_1

im_task_001160

HS_2
options: Prove theorems about triangles. Theorems include: measures of interior angles of a triangle sum to 180°; base angles of isosceles triangles are congruent; the segment joining midpoints of two sides of a triangle is parallel to the third side and half the length; the medians of a triangle meet at a point.

Prove theorems about lines and angles. Theorems include: vertical angles are congruent; when a transversal crosses parallel lines, alternate interior angles are congruent and corresponding angles are congruent; points on a perpendicular bisector of a line segment are exactly those equidistant from the segment’s endpoints.

Prove theorems about parallelograms. Theorems include: opposite sides are congruent, opposite angles are congruent, the diagonals of a parallelogram bisect each other, and conversely, rectangles are parallelograms with congruent diagonals.
problem_activity: Task
Below is an isosceles triangle $ABC$ with $|AB| = |AC|$:
[IMG]
Three students propose different arguments for why $m(\angle B) = m(\angle C)$.
Ravi says
If I draw the bisector of $\angle A$ then this is a line of symmetry for $\triangle ABC$ and so $m(\angle B) = m(\angle C)$.
Brittney says
If $M$ is the midpoint of $\overline{BC}$ then $\triangle ABM$ is congruent to $\triangle ACM$ and so $\angle B$ and $\angle C$ are congruent.
Courtney says
If $P$ is a point on $\overline{BC}$ such that $\overleftrightarrow{AP}$ is perpendicular to $\overline{BC}$ then $\triangle ABP$ is congruent to $\triangle ACP$ and so $\angle B$ and $\angle C$ are congruent.
Fill in the details in each argument to show why $m(\angle B) = m(\angle C)$. Can you find another different argument showing that $m(\angle B) = m(\angle C)$?


correct_option_index: 0