Question

the figure shows equilateral triangle abo with sides of length 10 in the xy-plane. segment ab is perpendicular to the x-axis. the terminal ray of an angle θ (not shown) in standard position passes through the point b. what is the value of $10\sin\theta$?
a $-5\sqrt{3}$
b $-\frac{\sqrt{3}}{2}$

Explanation:

Step1: Find coordinates of A, O, B

Let \(O=(0,0)\). Since \(ABO\) is equilateral, \(AB \perp x\)-axis. Let \(A=(x, y)\), \(B=(x, 0)\) (as \(AB\) is vertical). \(OA = 10\), \(AB=10\), so \(y=10\). For equilateral triangle, x-coordinate of \(A\): \(x = -\frac{10}{2} = -5\) (since \(AB\) is left of y-axis to have \(B\) with negative x). So \(B=(-5, 0)\)? No, wait: height of equilateral triangle is \(h = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3}\). Correct coordinates: \(O=(0,0)\), \(A=(-5, 5\sqrt{3})\), \(B=(-5, -5\sqrt{3})\) (since \(AB \perp x\)-axis, length 10, so y difference is 10).

Step2: Calculate \(\sin\theta\) for point B

For point \(B(x,y)=(-5, -5\sqrt{3})\), \(r = \sqrt{x^2 + y^2} = 10\) (distance from origin). \(\sin\theta = \frac{y}{r} = \frac{-5\sqrt{3}}{10}\)

Step3: Compute \(10\sin\theta\)

Multiply \(\sin\theta\) by 10: \(10 \times \frac{-5\sqrt{3}}{10}\)

Answer:

A. \(-5\sqrt{3}\)