Problem solving

Question

Finding the area of an isosceles triangle with the two angles being 38.5 degrees and the top angle being 103 degrees, with a side length of 35 feet or minutes(I'm unsure which one they are asking for). Do I need to convert to 35 minutes? should I just use the regular formula, a=1/2bhcos(x)?..I got 934.34...not sure if that is right or not

Answer 1

Area = (1/2)*base*height

 

Drop an altitude line from the 103^o angle to the base, forming two congruent right triangles.  The altitude line is the height of the triangle:

 

First, find the base:

• cos(38.5^o) = (base/2)/35 ft
• 35*cos(38.5^o) = (1/2)base

 

Now find the height:

• sin(38.5^o) = height/35 ft
• 35*sin(38.5) = height

 

Now compute the area:

• Area = (1/2)*base*height
• Area = 35*cos(38.5^o)*35*sin(38.5^o)

 

Use your calculator to compute the answer.