Free Account giveaway USEast/SC/L

[Dan]

sorry nexus, no dice

Exorcise won. pm'd acc. have fun.

mod can close this now

answer with work for anyone who cares.

a = ? = 13.03
b = 12
c = 10
A = 72
B = ? = 61.13
C = ? = 46.87
h = ? = 9.56

a^2 = 12^2 + 10^2 - 2*12*10*cos(72)
a^2 = 169.83
a = 13.03

sin©/10 = sin(72)/13.03
sin© = 10*sin(72)/13.03
sin© = 0.72
C = asin(0.72) = 46.87

B = 180 - 72 - 46.87 = 61.13

b = x+y = 10
y = 10 - x

h^2 = c^2 + x^2 - 2*c*x*cos(A)
h^2 = a^2 + y^2 - 2*a*y*cos©

c^2 + x^2 - 2*c*x*cos(A) = a^2 + y^2 - 2*a*y*cos©
10^2 + x^2 - 2*10*x*cos(72) = 13.03^2 + (10-x)^2 - 2*13.03*(10-x)*cos(46.87)
100 + x^2 - 6.18x = 169.78 + 100 - 20x + x^2 - 178.1 + 17.81x
-6.18x = 169.78 -178.1 - 20x + 17.81x
-6.18x = -8.32 - 2.19x
x = -8.32/-3.99
x = 2.085

y = 10 - 2.085 = 7.915

h^2 = 10^2 + 2.085^2 - 2*10*2.085*cos(72)
h = 9.56

h^2 = 13.03^2 + 7.915^2 - 2*13.03*7.915*cos(46.87)
h = 9.56



other guys answers
A = 72
B = 61.133
C = 46.867
a = 13.032
b = 12
c = 10
h = 9.511

Solution (if you even care to look at it):

Given: A=72 b=12 c=10

Law of Cosines to find a.

a^2=b^2 + c^2 - 2bc*cos(A)
a = 13.032

Law of Cosines to find B.
b^2=a^2 + c^2 - 2ac*cos(B)
B = 61.133

180 degrees in triangle to find C
180 = A + B + C
C = 46.867

Law of Sines and Simultaneous Eq for H.
Eq. 1 h/sin© = a/sin(180-x)
Eq. 2 h/sin(A) = c/sin(x)

With (180-x) + (x) being the two angles made where 'h' intersects 'b'.

Substituting h = [c*sin(A)] / sin(x) into Eq. 1
[c*sin(A)] / [sin(x)*sin©] = a/sin(180-x)

Simplifying
[c*sin(A)] / [a*sin©] = sin(x) / sin(180-x)

Substituting and Solving
1 = sin(x) / sin(180-x)
x = 90

Substituting x into Eq. 2 yields
h = 9.511


Now I'm just wondering how many of these kids actually pay attention in math.

damn shoulda said they were both wrong since the angle measurements are not in DMS