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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Answer 1 · 2017-04-26T01:31:20

$angleBCA=44^@$

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Let $angleABQ=QBC=y, and angleBAP=anglePAC=x$

In $DeltaADB, x-anglePAD+2y=90$
$x-6+2y=90^@$
$=> x+2y=96$ ---------------- EQ(1)

In $DeltaAEB, 2x+y-angleQBE=90$
$2x+y-18=90^@$
$=> 2x+y=108$ ---------------- EQ(2)

Multiplying EQ(1) by 2, we get
$2x+4y=192$ --------------- EQ(3)

Subtracting EQ(2) from EQ(3) yields:
$3y=84, => y=84/3=28$

Substituting $y=28$ into EQ(1), we get
$x+2xx28=96$
$=> x=40$

Hence, $angleBCA=180-(angleCAB+angleABC)$
$=180-2(x+y)$
$=180-2(40+28)=44^@$

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 angleangle PAD = 6° and angleangleQBE =18° , what is the degree of angleangleBCA?