The sine of 80 degrees is defined as the ratio of the opposite side to the hypotenuse, which here is 16/a.Hey, picture attached. Hoping that this can be solved via sin cos and tan but it’s been to long for me to remember that.
Thanks in advance,
NathanView attachment 28616
Honestly no. I’m trying to figure out the optimal angle for a specific build and I read that you needed that size for a strong build on the back. 20 years ago yes I learnt and remembered but I can’t now. Pythagorean no problem but the trig escapes meI took your statement to mean it's been too long to do it without a little help. We normally help people who want to learn something. But we can go further when needed.
But here's the work:
[imath]\sin(80^\circ) = \frac{16}{a}\\ a\sin(80^\circ) = 16\\ a = \frac{16}{\sin(80^\circ)} = \frac{16}{0.9848} \approx 16.25[/imath][imath]\tan(80^\circ) = \frac{16}{b}\\ b\tan(80^\circ) = 16\\ b = \frac{16}{\tan(80^\circ)} = \frac{16}{5.67} \approx 2.82[/imath]
Is that enough?