Confirmation

[ElPimpo]

Can u guys verify that this is correct.

If i have a triangle with sides 8, 11 and 17, would the length of the altitude drawn to the SMALLEST side be 10.99 or 11

/ \
8 / \11
/ ___ \
17


i used Sin and Cos to find out the measures of each of the angles which turned out to be

111.61
/ \
40.32____28.07

then to solve for the altitude from the vertex of 11 and 17 to 8 (which would be the smallest side)
i did

Sin 40.32 = alt/17 SOH(opposite/hypotenuse)
\ /
17 * Sin(40.32) = alt

and arrived at 10.99 or 11

would this be a way to solve it?

Thanks

[opus]

You have:
a=8
b=11
c=17

Using those formulas:

alpha = acos((a * a - b * b - c * c) / (-2 * b * c))
beta = acos((b * b - c * c - a * a) / (-2 * c * a))
gamma = acos((c * c - a * a - b * b) / (-2 * a * b))
ha = b * sin(gamma)
hb = c * sin(alpha)
hc = a * sin(beta)
I get:

alpha=22,31 degrees
beta=31,47 degrees
gamma=126,22 degrees
height on a (ha) =8,87
hb=6,45
hc=4,18

I didn't get any 40,32 degrees nor 10,99 height????

[sql_lall]

I don't have a calculator, but here goes:

Area of triangle = smallest side * altitude to smallest side
=> altitude to smallest side = Area/8

Area = sqrt(s*(s-a)*(s-b)*(s-c)) (Heron's, where s = (a+b+c)/2)

s = 36/2 = 18
a=8 b=11 c=17
=> Area = sqrt(18*10*7*1) = sqrt(1260) = 6 x sqrt(35)
=> altitude to smallest side = sqrt(35) * 6 / 8
Which is about 4.437 (4sf)

[opus]

Funny way to get a height,
that way you only get half the height!

Remember the area of a triangle is HALF of (Base * height on Base)

[sql_lall]

Yeah sorry.
I guess its the thought that counts