[Closed] A question for geometry lovers:

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Do I win a prize?

3.26

Do I win a prize?

Where are your workings?

Or in this case, maybe not.

Cut the triangle in half (to make two right angled triangles). It's just SOHCAHTOA then 😉

Take a midpoint of BC, call it D.

Draw a line between A and D. We now have two right angled triangles, with an angle A/2 of 50 and a known Opposite of 2.5.

Trig tells us SIN(angle) = O / H.

So, H = O / SIN(angle)

H = 2.5 / SIN(50)

H = 3.26

I think. It's a quarter of a century since I last did that.

Cougar wins!
Thank you......

Assuming he's right of course.

It feels about right.

The 100' angle will give you slightly shorter sides than if it were a 90' angle. If it were 90' then you could use Pythagoras' Theorymajig.

[AB]^2 + [AC]^2 = [BC]^2

As AB and AC are the same,

[BC]^2 = 2([AB]^2)

[AB]^2 = [BC]^2 / 2

[AB] = SQRT(25/2)

(ie, square the hypotenuse, divide by two, and square root it)

Which gives us a side of 3.53 - nominally longer than the 3.26 for the 100' angle, as predicted.

Yeah 3.26 is correct but there is no need to mess about with pythagoras. The internal angles are 100, 40, 40 and you know one length so it's just a matter of applying the sine rule.

Is there a second prize for getting the answer first?

Is there a prize for being first to provide instructions but not actually doing the workings for the lazy OP? 🙂

Cougar is correct.

there is no need to mess about with pythagoras.

Oh, I know, I was just doing it as a half-proof to see if I was in the right ballpark. Last time I did trig was at school so I wasn't sure I'd remembered it correctly.

It's got something to do with pie right? It's always about the Pie!

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😉