Question 453480
If we draw a height h from vertex C it divides side c in point D into two pieces x=AD and y=BD
:
Also it forms two right triangles:
- triangle ADC with one leg being height other one being x and b for hypotenuse
- triangle BDC with one leg being height other one being y and a for hypotenuse
:
We have for x, y and c:
x+y=c -> x+y=15 -> x=15-y
:
In triangle ADC we have {{{tan(42)=h/x}}} -> {{{h=x*tan(42)}}}
In triangle BDC we have {{{tan(68)=h/y}}} -> {{{h=y*tan(68)}}}
:
Now we have -> {{{x*tan(42)=y*tan(68)}}}
If we substitute x with x=15-y we have:
{{{(15-y)*tan(42)=y*tan(68)}}}
{{{15*tan(42)-y*tan(42)=y*tan(68)}}}
{{{15*tan(42)=y*tan(42)+y*tan(68)}}}
{{{y=15*tan(42)/(tan(42)+tan(68))}}}
:
From here:
{{{x=15-y}}}
{{{x=15*tan(68)/(tan(42)+tan(68))}}}
:
Knowing x we can find b=AC:
{{{cos(42)=x/b}}}
{{{b=x/cos(42)}}}
{{{b=15*tan(68)/cos(42)(tan(42)+tan(68))}}}
:
b=14.80
:
Knowing y we can find c=BC:
{{{cos(68)=y/c}}}
{{{c=y/cos(68)}}}
{{{c=15*tan(42)/cos(68)(tan(42)+tan(68))}}}
:
c=10.68