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
<em>Answer:</em>
<em>0 because 0 is sea level, anything below is negative and anything above is positive.</em>
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Answer:
6a - 9
Step-by-step explanation:
First off, let's put all of the identical variables besides each other, to make this equation easier.
-5ab - (-5ab) + 6a - 8 - 1
Then, we simply solve.
-5ab + 5ab = 0
6a = 6a
-8 - 1 = -9
So you're left with 6a - 9 as your answer.
Hope that helps!
Answer:
963 hours
Step-by-step explanation:
I subtracted 20 from the 981