SOLVE PLEASE. Would we use tan70=x/5.0?
1 answer:
I forgot my trig but I can see if I can do something
this will be a confusing solution
if we figure out the 3rd angle, it is 55
therefor we has a sideways isocoleese triangle (see attachment)
we can draw an auxilary (imaginary) line from left tower to left side to make 2 identical right triangles with hypotonuse 5 and the angles are 35 and 55
erase all that, and now we have
draw an immaginary line (see attachmen 2) from bottom to top to represent height
I am going to try to find the base (leg)
cos=a/h
cos(70°)=x/5
times both sides by 5
5cos(70°)=x (from left tower to perpendicular line)
now
a²+b²=c²
(5cos(70°)²+h²=5²
25(cos(70°))²+h²=25
minus 25(cos(70°))² both sides
h²=25-25(cos(70°))²
h²=25(1-(cos(70°)²))
square root both sides
h=5√(1-(cos(70°)²)) or 5cos(pi/9)
evaluate with your calculator
h=4.6984km
round
h=4.7km
this will be a confusing solution
if we figure out the 3rd angle, it is 55
therefor we has a sideways isocoleese triangle (see attachment)
we can draw an auxilary (imaginary) line from left tower to left side to make 2 identical right triangles with hypotonuse 5 and the angles are 35 and 55
erase all that, and now we have
draw an immaginary line (see attachmen 2) from bottom to top to represent height
I am going to try to find the base (leg)
cos=a/h
cos(70°)=x/5
times both sides by 5
5cos(70°)=x (from left tower to perpendicular line)
now
a²+b²=c²
(5cos(70°)²+h²=5²
25(cos(70°))²+h²=25
minus 25(cos(70°))² both sides
h²=25-25(cos(70°))²
h²=25(1-(cos(70°)²))
square root both sides
h=5√(1-(cos(70°)²)) or 5cos(pi/9)
evaluate with your calculator
h=4.6984km
round
h=4.7km
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