If $$ \cos \angle B=\sin \angle A $$ and $$ m \angle A=32^{\circ} $$, what is the measure of $$ \angle B $$ ?

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We are given:
$$
\cos \angle B = \sin \angle A \quad \text{and} \quad m \angle A = 32^\circ
$$
First, compute $$\sin \angle A$$:
$$
\sin 32^\circ
$$
We know the co-function identity:
$$
\sin \theta = \cos(90^\circ - \theta)
$$
Using this identity:
$$
\sin 32^\circ = \cos(90^\circ - 32^\circ) = \cos 58^\circ
$$
Therefore:
$$
\cos \angle B = \cos 58^\circ
$$
This implies:
$$
\angle B = 58^\circ \quad \text{or} \quad \angle B = 360^\circ - 58^\circ = 302^\circ
$$
However, in the context of typical geometrical problems (such as within a triangle), angles are usually between $$0^\circ$$ and $$180^\circ$$. Thus, the measure of $$\angle B$$ is:
$$
\boxed{58^\circ}
$$
The measure of angle B is 58 degrees.

If \( \cos \angle B=\sin \angle A \) and \( m \angle A=32^{\circ} \), what is the measure of \( \angle B \) ?

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