What is the pow of a 4.5 x 10-3 M OH solution?
1 answer:
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Answer:
The altitude of the climber is 1.40 miles
Step-by-step explanation:
see the attached figure to better understand the problem
we know that
<em>In the right triangle BCD</em>
tan(22°)=h/(x-0.5)
h=tan(22°)*(x-0.5) ----> equation A
<em>In the right triangle ACD</em>
tan(15°)=h/x
h=tan(15°)*(x) ----> equation B
Equate equation A and equation B and solve for x
tan(22°)*(x-0.5)=tan(15°)*(x)
tan(22°)*x-tan(22°)*0.5=tan(15°)*x
x[tan(22°)-tan(15°)]=tan(22°)*0.5
x=tan(22°)*0.5/[tan(22°)-tan(15°)]
x=1.48 miles
Find the value oh h
h=tan(15°)*(1.48)=0.40 miles
therefore
The altitude of the climber is equal to
0.40+1=1.40 miles
