1999 AHSME Problems/Problem 27

Problem

In triangle $ABC$ , $3 \sin A + 4 \cos B = 6$ and $4 \sin B + 3 \cos A = 1$ . Then $\angle C$ in degrees is

$\mathrm{(A) \ }30 \qquad \mathrm{(B) \ }60 \qquad \mathrm{(C) \ }90 \qquad \mathrm{(D) \ }120 \qquad \mathrm{(E) \ }150$

Solution

Square the given equations and add (simplifying with the Pythagorean identity $\sin^2 x + \cos^2 x = 1$ ):

$\begin{align*} 9\sin^2 A + 16\cos^2 B + 24 \sin A \cos B & = 36 \\ + 9\cos^2 A + 16\sin^2 B + 24 \sin B \cos A & = 1 \\ \Longrightarrow 25 + 24(\sin A \cos B + \sin B \cos A ) & = 37 \end{align*}$

Thus $\frac 12 = \sin A \cos B + \sin B \cos A$ . This is the sine addition identity, so $\frac 12 = \sin (A + B) = \sin (180 - C) = \sin C$ . Thus either $C = 30^{\circ}, 150^{\circ}$ .

If $C = 150$ , then $A + B = 30 \Longrightarrow A,B < 30$ , and $\sin A < \frac 12, \cos A < 1$ . The first equation implies $6 = 3 \sin A + 4\cos B < 3\left(\frac 12\right) + 4(1) = 5.5 < 6$ , which is a contradiction; thus $C = 30 \Longrightarrow \mathrm{(A)}$ .

Supplement

Adding the two given equations gives $3(\sin A + \cos A)+4(\sin B + \cos B) =7$

If $A+B = 30^{\circ}$ , then $A$ and $B$ are both acute. When $A$ and $B$ are both acute, $\sin A + \cos A > 1$ and $\sin B + \cos B>1$ . Then $3(\sin A + \cos A)+4(\sin B + \cos B) >7$ . This contradicts the equation $3(\sin A + \cos A)+4(\sin B + \cos B) =7$ . Therefore, $A+B \neq 30^{\circ}$ , so $A+B = 150^{\circ}$ and $C = 30^{\circ}$ .

~isabelchen

1999 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 26	Followed by Problem 28
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1999 AHSME Problems/Problem 27

Contents

Problem

Solution

Supplement

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