1984 AHSME Problems/Problem 23

Problem

$\frac{\sin{10^\circ}+\sin{20^\circ}}{\cos{10^\circ}+\cos{20^\circ}}$ equals

$\mathrm{(A) \ }\tan{10^\circ}+\tan{20^\circ} \qquad \mathrm{(B) \ }\tan{30^\circ} \qquad \mathrm{(C) \ } \frac{1}{2}(\tan{10^\circ}+\tan{20^\circ}) \qquad \mathrm{(D) \ }\tan{15^\circ} \qquad \mathrm{(E) \ } \frac{1}{4}\tan{60^\circ}$

Solution

From the sum-to-product formulas, we have $\sin{a}+\sin{b}=2\sin{\frac{a+b}{2}}\cos{\frac{a-b}{2}}$ . So, $\sin{10^\circ}+\sin{20^\circ}=2\sin{15^\circ}\cos{5^\circ}$ . Also from the sum-to-product formulas, we have $\cos{a}+\cos{b}=2\cos{\frac{a+b}{2}}\cos{\frac{a-b}{2}}$ , so $\cos{10^\circ}+\cos{20^\circ}=2\cos{15^\circ}\cos{5^\circ}$ . Substituting these back into the original fraction, we have $\frac{\sin{10^\circ}+\sin{20^\circ}}{\cos{10^\circ}+\cos{20^\circ}}=\frac{2\sin{15^\circ}\cos{5^\circ}}{2\cos{15^\circ}\cos{5^\circ}}=\frac{\sin{15^\circ}}{\cos{15^\circ}}=\tan{15^\circ}, \boxed{\text{D}}$ .

1984 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

1984 AHSME Problems/Problem 23

Problem

Solution

See Also