The distance between the points (a cos 25°, 0) and (0, a cos 65°) is

We have to find the distance between $\mathrm{A}\left(a \cos 25^{\circ}, 0\right)$ and $\mathrm{B}\left(0, a \cos 65^{\circ}\right)$.

In general, the distance between $\mathrm{A}\left(x_{1}, y_{1}\right)$ and $\mathrm{B}\left(x_{2}, y_{2}\right)$ is given by,

$\mathrm{AB}=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

So,

$A B=\sqrt{\left(0-a \cos 25^{\circ}\right)^{2}+\left(a \cos 65^{\circ}-0\right)^{2}}$

$=\sqrt{\left(a \cos 25^{\circ}\right)^{2}+\left(a \cos 65^{\circ}\right)^{2}}$

$\cos 25^{\circ}=\sin 65^{\circ}$ and $\cos 65^{\circ}=\sin 25^{\circ}$

But according to the trigonometric identity,

$\sin ^{2} \theta+\cos ^{2} \theta=1$

Therefore,

$\mathrm{AB}=a$

So, the answer is (a)