Question 86994
<pre>
Please help me solve this problem: the vertex 
angle of an isosceles triangle is 57°24' 
and each of its equal sides is  375.5 ft. long.
Find the altitude of the 
triangle.

<font size = 4 color = "darkblue"><b>
The solution by Stanbon contains an error in
calculation. He did it a different way, i.e.,
by calculating the base angles, but he calculated 
the base angles as 62°18' and they should have 
been 61°18'. 

A triangle has three altitudes.  I assume you 
mean the altitude from the vertex angle to the 
base. 
{{{drawing(400,350,-200,200,-10,340, triangle(-180.3239233,0,180.3239233,0,0,329.3683845), locate(-166,160,"375.5`ft"),
locate(120,160,"375.5`ft"), locate(-20,280,"57°24'")   
 ) }}}
Now draw in the altitude (call it "a") from the 
vertex angle to the base.

{{{drawing(400,350,-200,200,-10,340, triangle(-180.3239233,0,180.3239233,0,0,329.3683845), locate(-166,160,"375.5`ft"),
locate(110,160,"375.5`ft"), locate(-23,280,"57°24'"), line(0,0,0,329.3683845),
locate(8,160,"a"), rectangle(0,10,-10,0)

) }}}
 The altitude divides the triangle into two 
congruent right triangles, and divides the vertex 
angle by 2.  So each angle at the top is one-half of 
57°24'. Since 57° is an odd number, and doesn't divide
evenly by 2, borrow one degree from the 57°, 
making it an even number 56°, change the borrowed degree 
to 60' and add it to the 24', giving 56°84':

57°48 ÷ 2 = 56°84' ÷ 2 = 28°42' 

Now let's erase the right triangle on the right side 
and just look at the one on the left:

{{{drawing(400,350,-200,200,-10,340, triangle(- 180.3239233,0,0,0,0,329.3683845), locate(-166,160,"375.5`ft"), locate(-50,240,"28°42'"), line(0,0,0,329.3683845), locate(8,160,"a"), rectangle(0,10,-10,0)

) }}}

The altitude "a" is the side adjacent to the 28°42' angle.
The side which is 375.5 feet is the hypotenuse.  So you
need the basic trig function that involves adjacent and
hypotenuse, which is the cosine, so we have

cos(28°42') = {{{adjacent/hypotenuse}}}

cos(28°42') = {{{a/375.5}}}

Multiply both sides by 375.5

(375.5)cos(28°42') = a

The easiest way to handle 28°42' is to put the minutes
over 60 and add that to the number of degrees, that is, 
enter

375.5 × cos(28 + 42/60)

on your calculator to get

a = 329.3683845.  You should round
that to tenths since the given side
was rounded to tenths.  So the
altitude from the vertex angle to the 
base in the original isosceles triangle
is: 

a = 329.4 ft.

{{{drawing(400,350,-200,200,-10,340, triangle(-180.3239233,0,180.3239233,0,0,329.3683845), locate(-166,160,"375.5`ft"),
locate(110,160,"375.5`ft"), locate(-23,280,"57°24'"), line(0,0,0,329.3683845),
locate(8,150,"a=329.4`ft"), rectangle(0,10,-10,0) )}}}

Edwin</pre>