Solving for an angle using trigonometry.

Question

Trying again

$\dfrac{\sin(180 - \theta - a)}{H + R} = \dfrac{\sin(a)}{R}$

$\dfrac{\sin(\theta + a)}{ H + R} = \dfrac{\sin (a) }{ R}$ (is this correct?)

${\sin(\theta + a)} = \dfrac{{ H + R}}{ R}\sin (a) $

${(\theta + a)} = arcsin(\dfrac{{ H + R}}{ R}\sin (a))$

${\theta} = arcsin(\dfrac{{ H + R}}{ R}\sin (a)) - a$

allows me to get the arc x = ${\theta} * R$

Answer 1

from your last one:

$\sin{(\theta+a)}=\sin{\theta}\cos{a}+\cos{\theta}\cos{a}=x\cos{a}+\sqrt{1-x^2}\cos{a}$

$x=\sin{\theta}$

$x\cos{a}+\sqrt{1-x^2}\cos{a}=\dfrac{{ H + R}}{ R}\sin (a)$

solve $x$ ,you can get $\sin{\theta}$.(I suppose $a$ is known.)