if we assume the bottom right corner is a right angle and is the center of the arc, then it is solvable in the manners that others here have already described. if either of those is not the case, and the image itself doesn’t state, then there is insufficient information to solve it.
Draw line theta from left to right intersections and ya got two triangles.
Pythagoras gets you the length of that line. Trig gets you the two remaining angles of the red triangle (sohcahtoa!!)
180-angles gets you one angle of the new triangle.
You have to assume its an oblique triangle so make a perpendicular from x to get two right triangles, and then use some algebra on the lengths that you know and don’t know. Then trig again gets you cos theta = x/length and you get the remaining angles. Maybe I left a step out but generally thats the process.
if we assume the bottom right corner is a right angle and is the center of the arc, then it is solvable in the manners that others here have already described. if either of those is not the case, and the image itself doesn’t state, then there is insufficient information to solve it.
You sure?
Draw line theta from left to right intersections and ya got two triangles. Pythagoras gets you the length of that line. Trig gets you the two remaining angles of the red triangle (sohcahtoa!!) 180-angles gets you one angle of the new triangle.
You have to assume its an oblique triangle so make a perpendicular from x to get two right triangles, and then use some algebra on the lengths that you know and don’t know. Then trig again gets you cos theta = x/length and you get the remaining angles. Maybe I left a step out but generally thats the process.