Posted by carlo on Monday, March 19, 2012 at 7:25am.
In your sketch, draw an altitude hitting the base line at P, Howard to P is HP, Keight to P is KP
Let HP = x ,angle at H is 50°
then KP = 2-x, angle at K is 65
let the height be h
I see two right - angled triangles,
in the first: cot50 = x/h
x = hcot50
in the second: cot 65 = (2-x)/h
hcot65 = 2-x
x = 2 - hcot65
so
hcot50 = 2-hcot65
hcot50 + hcot65 = 2
h(cot50+cot65) = 2
h = 2/(cot50+cot65)
(only now do we have to go to a calculator, I will let you do the button-pushing)
for the second part, I see it as
cot50 = (2+x)/h and cot 65 = x/h
proceed the same way as above.
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