Since 9+9=19<20 and 10+10=20=20, the triangles with sides 9, 9, 20 and 10, 10, 20 cannot exist.
Use the cosine theorem to check which option C or D is true:
1.
\(20^2=14^2+14^2-2\cdot 14\cdot 14\cos \theta,\\ 400=196+196-392\cos \theta,\\ 8=-392\cos \theta,\\ \\ \cos \theta=-\dfrac{8}{392} =-\dfrac{1}{49}\).
You have that cosine of the greatest angle (opposite to the greatest side) is negative, this means that angle is obtuse.
2.
\(20^2=15^2+15^2-2\cdot 15\cdot 15\cos \theta,\\ 400=225+225-450\cos \theta,\\ -50=-450\cos \theta,\\ \\ \cos \theta=\dfrac{50}{450} =\dfrac{1}{9}\).
Here you have that cosine of the greatest angle (opposite to the greatest side) is posiative, this means that angle is acute.
Answer: the greatest possible whole-number value of the congruent side lengths is 14 (choice C)
