# In triangle ABC, AD and BE are altitudes and AP and BQ are angle bisectors at A and B respectively, where P lies on CD and Q lies on CE. If #angle# PAD = 6° and #angle#QBE =18° , what is the degree of #angle#BCA?

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In triangle ABC, AD and BE are altitudes and AP and BQ are angle bisectors at

A and B respectively, where P lies on CD and Q lies on CE. If #angle# PAD = 6°

and #angle# QBE =18° , what is the degree of #angle# BCA?

In triangle ABC, AD and BE are altitudes and AP and BQ are angle bisectors at

A and B respectively, where P lies on CD and Q lies on CE. If

and

Let

In

In

Multiplying EQ(1) by 2, we get

Subtracting EQ(2) from EQ(3) yields:

Substituting

Hence,

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To find the degree of angle ( BCA ), we can use the angle bisector theorem and some angle properties in triangles.

Given: ( \angle PAD = 6^\circ ) ( \angle QBE = 18^\circ )

Using the angle bisector theorem, we know that: [ \frac{AP}{PD} = \frac{AB}{BD} ] [ \frac{BQ}{QE} = \frac{AB}{BE} ]

From the given angles, we can find the other angles in the triangle:

Since ( AD ) is an altitude, ( \angle ADB = 90^\circ ). Similarly, ( \angle BEC = 90^\circ ).

Given that ( \angle PAD = 6^\circ ), ( \angle PAB = 84^\circ ) (since ( \angle BAP = \angle PAD )).

Given that ( \angle QBE = 18^\circ ), ( \angle QBA = 72^\circ ) (since ( \angle ABQ = \angle QBE )).

Using the fact that ( \angle PAB + \angle QBA = \angle BAC ): [ \angle BAC = 84^\circ + 72^\circ = 156^\circ ]

Finally, ( \angle BCA ) is supplementary to ( \angle BAC ): [ \angle BCA = 180^\circ - \angle BAC ] [ \angle BCA = 180^\circ - 156^\circ ] [ \angle BCA = 24^\circ ]

Thus, the degree of angle ( BCA ) is ( 24^\circ ).

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