Sagot :
I'm assuming you have or are taking trigonometry because that is how I solved this. First find the angle between the hypotenuse and the horizontal leg. To do this I took the inverse tangent of 5/2 (opposite/adjacent). Notice how I took the inverse tangent of the vertical leg/ horizontal leg. My calculator showed the answer to be about 68 degrees. Because of some angle laws (I'm not sure which ones, just trust me) that angle in each triangle on that line (the one between the horizontal component and the hypotenuse) must all have the same degree measure. Thus, test each answer by finding the inverse tangent of the vertical leg/horizontal leg. If the answer is about 68 degrees, that triangle could be on the line as well.
A. invtan(10/8)≈51
B. invtan(10/4)≈68
C. invtan(15/6)≈68
D. invtan(25/10)≈68
A. invtan(10/8)≈51
B. invtan(10/4)≈68
C. invtan(15/6)≈68
D. invtan(25/10)≈68