How do i get the angle of the geometric figure?

Question

I have this problem:

The only information that i have is that $2x + \angle BAC = 90$, but from here i can't get more. Any hint is appreciated.

Answer 1

Get out your big guns.

$\angle PAC = \angle AQP = 90 - 2x$

$\angle APQ = 180 - 2(90-2x) = 4x$.

$\angle BPQ = 180 - 4x$

$\angle BQP = 180 - x - (180 - 4x) = 3x$.

$\angle BQC = 180 - \angle AQP - \angle BQP$

And $\angle BQC = 90 - x$.

So $180 - (90-2x) - 3x = 90-x$ so ... oh, dang...

Well, now we do sides using law of sine.

$\frac {QC}{BQ} = \sin x$.

$\frac {BQ}{\sin 4x} = \frac {PQ}{\sin x}$

So that should do it $BQ = \frac {QC}{\sin x} = \frac {PQ\sin 4x}{\sin x}$ so

$QC = PQ \sin 4x$ so

$\sin 4x = \frac {QC}{PC} = \frac 12$.

$\arcsin \sin 4x = \arcsin \frac 12$

$4x = 60$

$x = 15$.

Answer 2

ΔAPQ is isosceles triangle. Let y = ∠PAQ, We got ∠PQA = y, ∠BPQ = 2y

ΔBCQ is right triangle, $QC = BQ \sin(x)$

Apply Law of sines to ΔBPQ:

$$ \frac{BQ}{\sin(2y)} = \frac{PQ}{\sin(x)} = \frac{2 QC}{\sin(x)} = 2 BQ $$ $$ \sin(2y) = \frac{1}{2} → 2y=30° \text{ or } 150° → y=15° \text{ or } 75°$$ $$ 2x+y= 90° → x= 7.5° \text{ or } 37.5° $$

Answer 3

Due to the right and isosceles triangles, $$\angle BPQ = 2\times (90-2x) = 180°-4x$$

Then, use the sine rule for $ ΔBPQ$,

$$ \frac{\sin (180°-4x)}{\sin x} = \frac{BQ}{PQ}= \frac{QC/\sin x}{2QC}$$

Therefore,

$$\sin 4x = \frac{1}{2}$$

which leads to

$$4x= 30° \space \text{or} \space 150°$$

Hence, there are two solutions for $x$:

$$ x_1=7.5°, \space \space x_2=37.5°$$

Answer 4

Let $QM$ be an altitude of $\Delta PQB$ and $M$ is placed between $P$ and $B$.

Thus, $$QM=QC=\frac{1}{2}PQ,$$ which gives $$\measuredangle QPM=30^{\circ}$$ and $$\measuredangle A=15^{\circ}.$$ Can you end it now?

The case when $P$ is placed between $M$ and $B$ is the same.