A right triangle has a hypotenuse of length 18 and an angle of 35°, with a side opposite this angle of length 4. A second right triangle also has an angle of 35°, with a hypotenuse of length 9. Determine the length of the side opposite the 35° angle on the second triangle.

Answer 1

Let $x$ be the length of the side opposite the $35^\circ$ angle on the second triangle. Then in this triangle, the side adjacent to the $35^\circ$ angle is $x \cos 35^\circ = x \left( \frac{\sqrt{10} + \sqrt{2}}{4} \right)$.

[asy]
unitsize (0.6 cm);

pair A, B, C, D, E, F;

A = (2,2*sqrt(3));
B = (0,0);
C = (8,0);
D = (2,0);
E = (D + reflect((A + D)/2,(B + C)/2)*(D))/2;

draw(A--B--C--cycle);
draw(D--E);

label("$35^\circ$", D + (0.8,0.2));
label("$x$", (E + D)/2, N);
label("$9$", (D + C)/2, S);
label("$4$", (A + E)/2, NW);
label("$18$", (A + B)/2, W);
[/asy]

Then by the Law of Cosines on this triangle,
\begin{align*}
(x \cos 35^\circ)^2 + 9^2 - 2 \cdot (x \cos 35^\circ) \cdot 9 \cdot \cos 35^\circ &= (x \sin 35^\circ)^2 \\
\Rightarrow \qquad 8x &= 120 \\
\Rightarrow \qquad x &= \boxed{15}.
\end{align*}