1997 AJHSME Problems/Problem 12

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Problem

$\angle 1 + \angle 2 = 180^\circ$

$\angle 3 = \angle 4$

Find $\angle 4.$

pair H,I,J,K,L;
H = (0,0); I = 10*dir(70); J = I + 10*dir(290); K = J + 5*dir(110); L = J + 5*dir(0);
draw(H--I--J--cycle);
draw(K--L--J);
draw(arc((0,0),,(1,0),CW)); label("$70^\circ$",dir(35),NE);
draw(arc(I,I+dir(250),I+dir(290),CCW)); label("$40^\circ$",I+1.25*dir(270),S);
label("$1$",J+0.25*dir(162.5),NW); label("$2$",J+0.25*dir(17.5),NE);
label("$3$",L+dir(162.5),WNW); label("$4$",K+dir(-52.5),SE);
 (Error compiling LaTeX. draw(arc((0,0),,(1,0),CW)); label("$70^\circ$",dir(35),NE);
               ^
95d9c9cbab8663c7f4f24f83898ce0fa39962883.asy: 9.16: syntax error
error: could not load module '95d9c9cbab8663c7f4f24f83898ce0fa39962883.asy')

$\text{(A)}\ 20^\circ \qquad \text{(B)}\ 25^\circ \qquad \text{(C)}\ 30^\circ \qquad \text{(D)}\ 35^\circ \qquad \text{(E)}\ 40^\circ$

Solution

Using the left triangle, we have:

$\angle 1 + 70 + 40 = 180$

$\angle 1 = 180 - 110$

$\angle 1 = 70$

Using the given fact that $\angle 1 + \angle 2 = 180$, we have $\angle 2 = 180 - 70 = 110$.

Finally, using the right triangle, and the fact that $\angle 3 = \angle 4$, we have:

$\angle 2 + \angle 3 + \angle 4 = 180$

$110 + \angle 3 + \angle 4 = 180$

$110 + \angle 4+ \angle 4 = 180$

$2\angle 4 = 70$

$\angle 4 = 35$

Thus, the answer is $\boxed{D}$

See also

1997 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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
All AJHSME/AMC 8 Problems and Solutions

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