Difference between revisions of "1961 AHSME Problems/Problem 8"

(Created page with 'Let the two base angles of a triangle be ''A'' and ''B'', with ''B'' larger than ''A''. The altitude to the base divides the vertex angle ''C'' into two parts, <math>C_1 and C_2<…')
 
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Let the two base angles of a triangle be ''A'' and ''B'', with ''B'' larger than ''A''. The altitude to the base divides the vertex angle ''C'' into two parts, <math>C_1 and C_2</math>, with <math>C_2</math> adjacent to side ''a''. Then:
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Let the two base angles of a triangle be ''A'' and ''B'', with ''B'' larger than ''A''. The altitude to the base divides the vertex angle ''C'' into two parts, <math>C_1</math> and <math>C_2</math>, with <math>C_2</math> adjacent to side ''a''. Then:
  
 
(A) <math>C_1+C_2=A+B</math>
 
(A) <math>C_1+C_2=A+B</math>

Revision as of 17:09, 1 October 2009

Let the two base angles of a triangle be A and B, with B larger than A. The altitude to the base divides the vertex angle C into two parts, $C_1$ and $C_2$, with $C_2$ adjacent to side a. Then:

(A) $C_1+C_2=A+B$

(B) $C_1-C_2=B-A$

(C) $C_1-C_2=A-B$

(D) $C_1+C_2=B-A$

(E) $C_1-C_2=A+B$

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