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 angleQBE =18° , what is the degree of angleBCA?

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 angleQBE =18° , what is the degree of angleBCA?

1 Answer
Apr 26, 2017

angleBCA=44^@BCA=44

Explanation:

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Let angleABQ=QBC=y, and angleBAP=anglePAC=xABQ=QBC=y,andBAP=PAC=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^@