MHB Problems involving Trigonometric Identities

[bearn]

What are the step-by-step in solving these problems?

 

[HOI]

For (1) I assume the lower left angle is a right angle. Then we have two right triangles. For the smaller right triangle, $tan(\theta)= \frac{5}{x}$. For the larger, $tan(\theta+ \alpha)= tan(2\theta)= \frac{5+ 8}{x}= \frac{13}{x}$.

For $tan(2\theta)$ you can use the identity $tan(2\theta)= \frac{2 tan(\theta)}{1- tan^2(\theta)}$.

Since the first equation says that $tan(\theta)= \frac{5}{x}$, the second says $\frac{\frac{10}{x}}{1- \frac{25}{x^2}}= \frac{13}{x}$ that you can solve for x.

 

[HOI]

For (2) the blue line is given by 6x+ 7y= 60 or y= -(6/7)x+ 60/7. Its slope is -6/7 so the "exterior angle" of that triangle is arctan(-6/7)= 139.4 degrees. The interior angle is 180- 139.4= 40.6 degrees. Since $\theta= 60$ degrees the third angle, where the red line crosses the base is 180- 60- 40.6= 120- 40.6= 79.4 degrees. So the slope of the red line is tan(79.4)= 5.34.
y= 5.34(x- x_0)+ y_0 where (x_0, y_0) is any point on the line. We are told that (3, 6) is such a point.

 

[skeeter]

For (1) I assume the lower left angle is a right angle.

If not, the problem becomes quite messy ... 3 unknown sides & one unknown angle. I was able to come up with four equations, but I'll be damned if I would make a "by hand" attempt at solving them for x, y, & z.

 

[bearn]

For (2) the blue line is given by 6x+ 7y= 60 or y= -(6/7)x+ 60/7. Its slope is -6/7 so the "exterior angle" of that triangle is arctan(-6/7)= 139.4 degrees. The interior angle is 180- 139.4= 40.6 degrees. Since $\theta= 60$ degrees the third angle, where the red line crosses the base is 180- 60- 40.6= 120- 40.6= 79.4 degrees. So the slope of the red line is tan(79.4)= 5.34.
y= 5.34(x- x_0)+ y_0 where (x_0, y_0) is any point on the line. We are told that (3, 6) is such a point.

Got it! Thank You so much!