heron's formula is:
a = sqrt(s(s-a)(s-b)(s-c)) where:
a = area
s = (a + b + c)/2
s can also be shown as p/2 where:
p = a + b + c
in your triangle:
a = 36
b = 54
c = 77
using heron's formula, we get:
p = a + b + c = 167
p/2 = 167/2 = 83.5
a = sqrt(s(s-a)*(s-b)*s-c)) becomes:
a = sqrt(83.5(47.5)(29.5)(6.5)) which becomes:
a = 872.0828157
here's a solver from one of the folks at algebra.com that give you the answer with a lot less effort on your part.
http://www.algebra.com/algebra/homework/Triangles/Find-the-area-of-triangle-if-3-sides-are-given.solver
it's good to know the formula though.
on a test, you won't have access to the nifty little tool developed here.
best thing is to do the calculations yourself and then use the tool to check your work.
you could also have solved for the area using the law of cosines but it's a lot more work.
kudos to heron.
your second problem is different because you are given an angle.
assume your triangle is labeled ABC.
The angle that is 50 degrees would be A.
The side that is 99 cm in length would be AB.
The side that is 112 cm in length would be AC.
BC is the side opposite angle A, the length of which is unknown.
you want to find the length of BC which is the distance between the butts of each rifle.
you can use the law of cosines here.
re-label your triangle as follows:
AB = c
AC = b
BC = a
AB is opposite angle C which means that c is opposite angle C
AC is opposite angle B which means that b is opposite angle B
BC is opposite angle A which means that a is opposite angle A
the law of cosines states:
a^2 = b^2 + c^2 - 2*b*c*cos(A)
substitute known values into this formula to get:
a^2 = 112^2 + 99^2 - 2*112*99*cos(50) which becomes:
a^2 = 12544 + 9801 - 22176*.64278761 which becomes:
a^2 = 8090.541968
take square root of both sides of this equation to get:
a = 89.94744003
your triangles look like this:
