Asked by Edward
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.
And please show working out to help me understand.
Answers
Answered by
Steve
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.
OR--------
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.
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.
OR--------
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.
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