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

##### 1 Answer
Apr 26, 2017

$\angle B C A = {44}^{\circ}$

#### Explanation:

Let $\angle A B Q = Q B C = y , \mathmr{and} \angle B A P = \angle P A C = x$

In $\Delta A D B , x - \angle P A D + 2 y = 90$
$x - 6 + 2 y = {90}^{\circ}$
$\implies x + 2 y = 96$ ---------------- EQ(1)

In $\Delta A E B , 2 x + y - \angle Q B E = 90$
$2 x + y - 18 = {90}^{\circ}$
$\implies 2 x + y = 108$ ---------------- EQ(2)

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

Subtracting EQ(2) from EQ(3) yields:
$3 y = 84 , \implies y = \frac{84}{3} = 28$

Substituting $y = 28$ into EQ(1), we get
$x + 2 \times 28 = 96$
$\implies x = 40$

Hence, $\angle B C A = 180 - \left(\angle C A B + \angle A B C\right)$
$= 180 - 2 \left(x + y\right)$
$= 180 - 2 \left(40 + 28\right) = {44}^{\circ}$