Question

Points P, Q, and S are collinear. If ray QT bisects angle RQS, what is m angle TQS?

Answer 1

Answer:

The angle TQS has a value of 23.5º.

Step-by-step explanation:

$\angle RQS$ and $\angle PQR$ are supplementary to each other, that is, the sum of angles equals 180º, and ray $QT$ divides $\angle RQS$ into two equal parts. We construct this system of linear equations after a careful reading of statement:

$\angle RQS = (x+1)^{\circ}$ (Eq. 1)

$\angle PQR = (3\cdot x - 5)^{\circ}$ (Eq. 2)

$\angle RQS + \angle PQR = 180^{\circ}$ (Eq. 3)

$\angle RQS = \angle RQT + \angle TQS$ (Eq. 4)

$\angle RQT = \angle TQS$ (Eq. 5)

Now we proceed to solve the system algebraically:

(Eq. 1) and (Eq. 2) in (Eq. 3)

$(x+1)^{\circ} + (3\cdot x - 5)^{\circ} = 180^{\circ}$

$(4\cdot x-4)^{\circ} = 180^{\circ}$

$(4\cdot x)^{\circ} = 184^{\circ}$

$x = \frac{184}{4}$

$x = 46$

In (Eq. 1):

$\angle RQS = 47^{\circ}$

(Eq. 5) in (Eq. 4):

$\angle RQS = 2\cdot \angle TQS$

$\angle TQS = \frac{1}{2}\cdot \angle RQS$

$\angle TQS = \frac{1}{2}\cdot (47^{\circ})$

$\angle TQS = 23.5^{\circ}$

The angle TQS has a value of 23.5º.