# 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 $C = 30^{\circ}$.

## See also

 1999 AHSME (Problems • Answer Key • Resources) Preceded byProblem 26 Followed byProblem 28 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 All AHSME Problems and Solutions

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