1985 AHSME Problems/Problem 28

Problem

In $\triangle ABC$ , we have $\angle C=3\angle A, a=27,$ and $c=48$ . What is $b$ ?

$[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair A=origin, B=(14,0), C=(10,6); draw(A--B--C--cycle); label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, N); label("$a$", B--C, dir(B--C)*dir(-90)); label("$b$", A--C, dir(C--A)*dir(-90)); label("$c$", A--B, dir(A--B)*dir(-90));[/asy]$

$\mathrm{(A)\ } 33 \qquad \mathrm{(B) \ }35 \qquad \mathrm{(C) \ } 37 \qquad \mathrm{(D) \ } 39 \qquad \mathrm{(E) \ }\text{not uniquely determined}$

Solution

From the Law of Sines, we have $\frac{\sin(A)}{a}=\frac{\sin(C)}{c}$ , or $\frac{\sin(A)}{27}=\frac{\sin(3A)}{48}\implies 9\sin(3A)=16\sin(A)$ .

We now need to find an identity relating $\sin(3A)$ and $\sin(A)$ . We have

$\sin(3A)=\sin(2A+A)=\sin(2A)\cos(A)+\cos(2A)\sin(A)$

$=2\sin(A)\cos^2(A)+\sin(A)\cos^2(A)-\sin^3(A)=3\sin(A)\cos^2(A)-\sin^3(A)$

$=3\sin(A)(1-\sin^2(A))-\sin^3(A)=3\sin(A)-4\sin^3(A)$ .

Thus we have $9(3\sin(A)-4\sin^3(A))=27\sin(A)-36\sin^3(A)=16\sin(A)$

$\implies 36\sin^3(A)=11\sin(A)$ .

Therefore, $\sin(A)=0, \frac{\sqrt{11}}{6},$ or $-\frac{\sqrt{11}}{6}$ . Notice that we must have $0^\circ<A<45^\circ$ because otherwise $A+3A>180^\circ$ . We can therefore disregard $\sin(A)=0$ because then $A=0$ and also we can disregard $\sin(A)=-\frac{\sqrt{11}}{6}}$ (Error compiling LaTeX. Unknown error_msg) because then $A$ would be in the third or fourth quadrants, much greater than the desired range.

Therefore, $\sin(A)=\frac{\sqrt{11}}{6}$ , and $\cos(A)=\sqrt{1-\left(\frac{\sqrt{11}}{6}\right)^2}=\frac{5}{6}$ . Going back to the Law of Sines, we have $\frac{\sin(A)}{27}=\frac{\sin(B)}{b}=\frac{\sin(\pi-3A-A)}{b}=\frac{\sin(4A)}{b}$ .

We now need to find $\sin(4A)$ .

$\sin(4A)=\sin(2\cdot2A)=2\sin(2A)\cos(2A)$

$=2(2\sin(A)\cos(A))(\cos^2(A)-\sin^2(A))$

$=2\cdot2\cdot\frac{\sqrt{11}}{6}\cdot\frac{5}{6}\left(\left(\frac{5}{6}\right)^2-\left(\frac{\sqrt{11}}{6}\right)^2\right)=\frac{35\sqrt{11}}{162}$ .

Therefore, $\frac{\frac{\sqrt{11}}{6}}{27}=\frac{\frac{35\sqrt{11}}{162}}{b}\implies b=\frac{6\cdot27\cdot35}{162}=35, \boxed{\text{B}}$ .

1985 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 27	Followed by Problem 29
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1985 AHSME Problems/Problem 28

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