Can someone help me solve the following diagram, if at all possible...
Solve for Angle: Theta
Regards,
- Mark
If we call the hypotenuse of the smaller triangle , then our large triangle has dimensions (by Pythagoras).
Since the triangles are similar, their corresponding sides are in the same proportion, so
Therefore, the height of the larger triangle is , and so we have the relationship
Here's an alternative method.
First some construction. (Apologies, I don't know how to incorporate diagrams, maybe one of the sites resident experts could suggest a method ?)
Run a line from the centre of the circle parallel with the hypotenuse down to the base of the triangle. Call the three points along the base D, E and F. (D is at the end of the hypotenuse, E is the new point and F is at the rightangle). Run another line from E upto the hypotenuse to meet it at rightangles (it will be parallel with the line in the circle), and call this point G.
In the new triangle DEG, , but so
in which case
Therefore,
Mutiply top and bottom of the LHS by to get
cancel the 's, cross multiply, rearrange, and you arrive at
From hereon it's routine, put the LHS equal to expand and equate with the existing LHS to get
so that and .
Finally then,
S*** what an idiot !!!
Draw a line from the centre of the circle down to the base of the hypotenuse. The "new" hypotenuse will be of length .
The angle will be the sum of those two angles which will be, from the two triangles and
Draw the segment containing the two right angles. We'll call it PQ, where P is the right angle in the smaller triangle and Q is the right angle in the larger triangle. Suppose the three vertices of the large right triangle are, in clockwise order, QRS, where Q is a right angle.
The hypotenuse of the large triangle is given by . Also, .
Therefore,
Applying the law of cosines,
Equate these two, noting that
And so on.