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1988 AJHSME Problems/Problem 5

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Problem

If $\angle \text{CBD}$ is a right angle, then this protractor indicates that the measure of $\angle \text{ABC}$ is approximately

$\text{(A)}\ 20^\circ \qquad \text{(B)}\ 40^\circ \qquad \text{(C)}\ 50^\circ \qquad \text{(D)}\ 70^\circ \qquad \text{(E)}\ 1...$

Solution

We have that $20^{\circ}+\angle ABD +\angle CBD=160^{\circ}$ , or $\angle ABD +\angle CBD=140^{\circ}$ . Since $\angle CBD$ is a right angle, we have $\angle ABD=140^{\circ}-90^{\circ}=50^{\circ}\Rightarrow \mathrm{(C)}$ .

See Also

1988 AJHSME (Problems • Resources)
Preceded by Problem 4	Followed by Problem 6
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

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/1988_AJHSME_Problems/Problem_5"

Category: Introductory Geometry Problems

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