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Difference between revisions of "1997 AJHSME Problems/Problem 12"
 
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label("$3$",L+dir(162.5),WNW); label("$4$",K+dir(-52.5),SE);
 
label("$3$",L+dir(162.5),WNW); label("$4$",K+dir(-52.5),SE);
 
</asy>
 
</asy>
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<math>\text{(A)}\ 20^\circ \qquad \text{(B)}\ 25^\circ \qquad \text{(C)}\ 30^\circ \qquad \text{(D)}\ 35^\circ \qquad \text{(E)}\ 40^\circ</math>
 
<math>\text{(A)}\ 20^\circ \qquad \text{(B)}\ 25^\circ \qquad \text{(C)}\ 30^\circ \qquad \text{(D)}\ 35^\circ \qquad \text{(E)}\ 40^\circ</math>
  

Latest revision as of 11:37, 2 May 2021

Problem

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

$\angle 3 = \angle 4$

Find $\angle 4.$

[asy] 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),dir(70),(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); [/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 (Problems • Answer Key • Resources)
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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