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202305171045 Problem 23.11

In the figure below, AC is a diameter of the circle.

If AC=1, which of the following gives the area of triangle ABC in terms of \theta?

A. \displaystyle{\frac{\theta}{2}}
B. \displaystyle{\frac{\tan\theta}{2}}
C. 2\sin\theta
D. \displaystyle{\frac{\sin\theta\cos\theta}{2}}

Extracted from Phu Nielson. (2015). SAT Math Advanced Guide and Workbook.

Remark.

The quantity angle \theta is dimensionless as it is the ratio of arc length to radius, i.e.,

\frac{[L]}{[L]}=[L]^0;

the sine, the cosine, and the tangent of which, as in a right-angled triangle, are

\begin{aligned} \sin\theta & = \frac{\textrm{opposite}}{\textrm{hypotenuse}} \\ \cos\theta & = \frac{\textrm{adjacent}}{\textrm{hypotenuse}} \\ \tan\theta & = \frac{\textrm{opposite}}{\textrm{adjacent}} \\ \end{aligned}

also dimensionless, i.e., [L]^0. Note that the quantity area is [L]^2 in dimension. Hence, choices A. to D. are being understood as:

A. \displaystyle{\frac{\theta}{2}}\enspace\textrm{(sq unit)}
B. \displaystyle{\frac{\tan\theta}{2}}\enspace\textrm{(sq unit)}
C. 2\sin\theta\enspace\textrm{(sq unit)}
D. \displaystyle{\frac{\sin\theta\cos\theta}{2}}\enspace\textrm{(sq unit)}

Warm-up.

Set-up.

where

\begin{aligned} \mathbf{OA} & = (-1/2,0) \\ \mathbf{OB} & = (x,y) \\ \mathbf{OC} & = (1/2,0) \\ \mathbf{AB} & = \mathbf{OB}-\mathbf{OA} \\ & = (x+1/2,y) \\ AB &= |\mathbf{AB}| \\ & = \sqrt{\bigg( x+\frac{1}{2}\bigg)^2+y^2} \\ \mathbf{BC} & = \mathbf{OC}-\mathbf{OB} \\ & = (1/2-x,-y) \\ BC &= |\mathbf{BC}| \\ & = \sqrt{\bigg(\frac{1}{2}-x\bigg)^2+(-y)^2} \\ \mathbf{AC} & = \mathbf{OC}-\mathbf{OA} \\ & = (1,0) \\ AC & = |\mathbf{AC}| \\ & = \sqrt{(1)^2+(0)^2} =1\\ \sin\theta & = \frac{y}{BC} \\ \cos\theta & = \frac{1/2-x}{BC} \\ \tan\theta & = \frac{y}{1/2-x} \\ \end{aligned}

Observe that

\displaystyle{\textrm{Area of }\triangle ABC=\frac{|\mathbf{AB}||\mathbf{BC}|}{2}}

proceed with

\begin{aligned} &\quad \frac{AB\cdot BC}{2} \\ & = \frac{\sqrt{(x+0.5)^2+y^2}\sqrt{(0.5-x)^2+(-y)^2}}{2} \\ & = \frac{\sqrt{x^2+x+\frac{1}{4}+y^2}\sqrt{x^2-x+\frac{1}{4}+y^2}}{2} \\ & = \frac{\sqrt{\frac{1}{2}+x}\sqrt{\frac{1}{2}-x}}{2} \\ & = \frac{\sqrt{(\frac{1}{2})^2-x^2}}{2}\\ \end{aligned}

by noting OB=|\mathbf{OB}| the radius being \frac{1}{2}:

\displaystyle{x^2+y^2=\bigg(\frac{1}{2}\bigg)^2=\frac{1}{4}}.

The function

\displaystyle{f(x)=\sqrt{\frac{1}{16}-\frac{x^2}{4}}}

will output the area by inputting x under constraint on x, y:

g(x,y)=x^2+y^2-\frac{1}{4}=0.

(to be refreshed)

Now ready for problem-solving, write simply

\begin{aligned} \sin\theta & = \frac{AB}{AC}=AB\\ \cos\theta & = \frac{BC}{AC}=BC\\ \tan\theta & = \frac{AB}{BC}\\ \end{aligned}

and the answer is D.