Question

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 `angle` PAD = 6°
and `angle`QBE =18° , what is the degree of `angle`BCA?

Answer 1

`angleBCA=44^@`

Explanation:

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^@`