MATHS GEOMETRY!!! URGENT PLEASE.
posted by Edward on .
A triangle has side lengths 10,17 and 21cm. Find the length of the shortest altitude.
And please show working out to help me understand.
if a,b,c = 10,17,21,
a^2 = b^2+c^2-2bc cos C, so
cosC = (17^2 + 21^2 - 10^2)/(2*17*21) = 15/17
so, sinC = 0.47
figure sinA and sinB the same way
If the altitudes from A,B,C are Ha,Hb,Hc, then
Hc = a*sinB = b*sinA
Figure the other altitudes in like wise.
figure the area using Heron's formula.
Then since 1/2 c*Hc = area
Hc = 2*area/c
Since the area does not change, the shortest altitude is on the longest side.