- #1
Albert1
- 1,221
- 0
$\triangle ABC ,\angle C=90^o, \overline{AB}=c, \overline{BC}=a, \overline{CA}=b, x=\dfrac{a}{c}, y=\dfrac{b}{c}$
and satisfying: $13xy=15(x+y)-15,$ find $x+y$
and satisfying: $13xy=15(x+y)-15,$ find $x+y$
Albert said:$\triangle ABC ,\angle C=90^o, \overline{AB}=c, \overline{BC}=a, \overline{CA}=b, x=\dfrac{a}{c}, y=\dfrac{b}{c}$
and satisfying: $13xy=15(x+y)-15,$ find $x+y$
johng said:Albert,
This is a challenge for high school students?
Clearly $x^2+y^2=1$ and $13xy=15(x+y)-15$ or easily
$$13(x+y)^2=30(x+y)-17$$
Hence by the quadratic formula $x+y=1$ or $x+y={17\over13}$
The formula for solving for x+y in a triangle is x+y = c, where c is the length of the third side of the triangle.
You will need the lengths of two sides of the triangle and the measure of the included angle between those two sides to solve for x+y.
No, you must use the two sides that are connected by the included angle. Using any other sides will not give you an accurate answer for x+y.
If you only know the length of one side and the measure of the included angle, you can use the Law of Cosines to find the length of the other side. Then, you can use the formula x+y = c to solve for x+y.
Yes, if the triangle is a right triangle, you can use the Pythagorean Theorem to find the length of the third side. Then, you can use the formula x+y = c to solve for x+y.