1994 AJHSME Problems/Problem 7

Problem

If $\angle A = 60^\circ$ , $\angle E = 40^\circ$ and $\angle C = 30^\circ$ , then $\angle BDC =$

$[asy] pair A,B,C,D,EE; A = origin; B = (2,0); C = (5,0); EE = (1.5,3); D = (1.75,1.5); draw(A--C--D); draw(B--EE--A); dot(A); dot(B); dot(C); dot(D); dot(EE); label("$A$",A,SW); label("$B$",B,S); label("$C$",C,SE); label("$D$",D,NE); label("$E$",EE,N); [/asy]$

$\text{(A)}\ 40^\circ \qquad \text{(B)}\ 50^\circ \qquad \text{(C)}\ 60^\circ \qquad \text{(D)}\ 70^\circ \qquad \text{(E)}\ 80^\circ$

Solution

The sum of the angles in a triangle is $180^\circ$ . We can find $\angle ABE = 80^\circ$ , so $\angle CBD = 180-80=100^\circ$ .

$\angle BDC = 180-100-30=\boxed{\text{(B)}\ 50^\circ}$

1994 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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1994 AJHSME Problems/Problem 7

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