Question 982458

<B>1</B>. &nbsp;First, &nbsp;let us assume that the altitude of &nbsp;20 cm&nbsp; long goes&nbsp; to the base &nbsp;{{{b}}}&nbsp; and the altitude of &nbsp;30 cm&nbsp; long goes to the lateral side &nbsp;{{{a}}}.


Then you have these expressions for the area of the triangle:

{{{S}}} = {{{1/2}}}.{{{b*20}}} = {{{1/2}}}.{{{a*30}}}.


It implies that &nbsp;{{{b*20}}} = {{{a*30}}}.


Hence,  &nbsp;{{{b}}} = {{{a*(30/20)}}} = {{{a*(3/2)}}}.


Now consider the right-angled triangle formed by the base, &nbsp;by the altitude of &nbsp;20 cm&nbsp; long and by the lateral side &nbsp;{{{a}}}.


This right-angled triangle has the hypotenuse of the length &nbsp;{{{a}}}&nbsp; and the leg of the length &nbsp;{{{b/2}}}&nbsp; (yes, &nbsp;it is half of the base of the triangle, &nbsp;{{{b/2}}} = {{{a*(3/4)}}}). 


I believe &nbsp;(I hope)&nbsp; you just made a sketch of your triangle. &nbsp;It will help you to see what I am saying.


Therefore, &nbsp;{{{cos(alpha)}}} = {{{(b/2)/a}}} = {{{(a*(3/4))/a}}} = {{{3/4}}}, &nbsp;where &nbsp;{{{alpha}}}&nbsp; is the angle at the base of our isosceles triangle. 


Thus, &nbsp;in this case {{{cos(alpha)}}} = {{{3/4}}} and {{{alpha}}} = {{{arccos(3/4)}}}. 



<B>2</B>. &nbsp;Next, let us consider the other configuration when the altitude of &nbsp;30 cm&nbsp; long goes&nbsp; to the base &nbsp;{{{b}}}&nbsp; and the altitude of &nbsp;20 cm&nbsp; long goes to the lateral side &nbsp;{{{a}}}.


Then you have these expressions for the area of the triangle:

{{{S}}} = {{{1/2}}}.{{{b*30}}} = {{{1/2}}}.{{{a*20}}}.


It implies that &nbsp;{{{b*30}}} = {{{a*20}}}.


Hence,  &nbsp;{{{b}}} = {{{a*(20/30)}}} = {{{a*(2/3)}}}.


Now consider the right-angled triangle formed by the base, &nbsp;by the altitude of &nbsp;30 cm&nbsp; long and by the lateral side &nbsp;{{{a}}}.


This right-angled triangle has the hypotenuse of the length &nbsp;{{{a}}}&nbsp; and the leg of the length &nbsp;{{{b/2}}}&nbsp; (which is half of the base of the triangle, {{{b/2}}} = {{{a*(2/6)}}}) = {{{a*(1/3)}}}. 


Again, I hope you just made a sketch of your triangle. 


Therefore, &nbsp;{{{cos(alpha)}}} = {{{(b/2)/a}}} = {{{(a*(1/3))/a}}} = {{{1/3}}}. 


Thus, &nbsp;in this case {{{cos(alpha)}}} = {{{1/3}}} and {{{alpha}}} = {{{arccos(1/3)}}}. 


<B>Answer</B>. &nbsp;{{{alpha}}} = {{{arccos(1/3)}}}&nbsp; or &nbsp;{{{alpha}}} = {{{arccos(3/4)}}}.