Question 86385
<pre><font size = 3><b>
In an isoceles triangle each of the equal sides is 10 meters and the base is 16
meters. Find the vertex angle accurate to the nearest minute. 

{{{drawing(400,177.78,-9,9,-1,7, triangle(-8,0,8,0,0,6), 
locate(4,3.5,"10"), locate(-4,4,"10"), locate(0,0,"16")  
)}}}

Draw an altitude to the base. This altitude is also a median, as well
as an angle bisector.  So it splits the base into two 8-inch parts, as
weel as bisects the vertex angle:

{{{drawing(400,177.78,-9,9,-1,7, triangle(-8,0,8,0,0,6), 
locate(4,3.5,"10"), locate(-4,4,"10"), locate(-4,0,"8"), locate(4,0,"8"),
 line(0,0,0,6)  
)}}}

Just look at the laft half of the triangle.  It is a right triangle:

{{{drawing(400,177.78,-9,9,-1,7, triangle(-8,0,0,0,0,6), locate(-4,4,"10"), locate(-4,0,"8"), locate(0,6.7,"A"), 
 line(0,0,0,6) )}}}

sin(A) = {{{opposite/hypotenuse}}} = {{{8/10}}} = {{{.8}}}

A = 53°8'

Therefore the entire vertex angle is twice that value:

{{{drawing(400,177.78,-9,9,-1,7, triangle(-8,0,8,0,0,6), 
locate(4,3.5,"10"), locate(-4,4,"10"), locate(0,0,"16")  
)}}}

Vertex angle = 106°16'

Edwin</pre>