Difference between revisions of "1988 AJHSME Problems/Problem 5"
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<math>\text{(A)}\ 20^\circ \qquad \text{(B)}\ 40^\circ \qquad \text{(C)}\ 50^\circ \qquad \text{(D)}\ 70^\circ \qquad \text{(E)}\ 120^\circ</math> | <math>\text{(A)}\ 20^\circ \qquad \text{(B)}\ 40^\circ \qquad \text{(C)}\ 50^\circ \qquad \text{(D)}\ 70^\circ \qquad \text{(E)}\ 120^\circ</math> | ||
==Solution== | ==Solution== | ||
− | We have that <math>20^{\circ}+\angle ABD +\angle CBD=160^{\circ}</math>, or <math>\angle ABD +\angle CBD=140^{\circ}</math>. Since <math>\angle CBD</math> is a right angle, we have <math>\angle ABD=140^{\circ}-90^{\circ}=50^{\circ}\Rightarrow \mathrm{(C)}</math>. | + | We have that <math>20^{\circ}+\angle ABD +\angle CBD=160^{\circ}</math>, or <math>\angle ABD +\angle CBD=140^{\circ}</math>. Since <math>\angle CBD</math> is a right angle, we have <math>\angle ABD=140^{\circ}-90^{\circ}=50^{\circ}\Rightarrow \boxed{\mathrm{(C)}}</math>. |
==See Also== | ==See Also== |
Latest revision as of 15:53, 17 December 2020
Problem
If is a right angle, then this protractor indicates that the measure of is approximately
Solution
We have that , or . Since is a right angle, we have .
See Also
1988 AJHSME (Problems • Answer Key • 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 | ||
All AJHSME/AMC 8 Problems and Solutions |
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