Items 2-4. Points P, Q, and Sare collinear.
(3x - 5)
0
(x + 1)
s
2. What is the value of x?
3. What is m ZPQR?
will
11. M
4. If a ray QT bisects ZRQS, what
will be the measure of one of the
resulting angles?
m
of
5. Points L, M, and N are collinear.
You are given LM = 5 and LN = 17.
What is a possible value of MN?
Flb
6. Ray BD bisects ZABC so that
mZDBC = x + 6 and
mZABD = 2x - 12. What is x?

See below

Step-by-step explanation:

2

∠PQR and ∠RQS are supplementary, which add up 180°

Sum:

3

4

From the picture and the value of x we got above:

QT bisects ∠RQS and we get two equal angles

Each will measure:

5

LM = 5, LN = 17, MN = ?

If L is between M and N, then

M is between L and N, then

5

Ray BD bisects ∠ABC, then formed angles are equal:

m∠DBC = m∠ABD

hope this !

step-by-step explanation:

d = distance between moon and sun, which in this problem is the hypotenuse of a right triangle.  

the distance y is the side opposite to angle x in the right triangle.   use the definition of the sine function from basic trigonometry.

sin(x) = y/d ⇒

d = y/sin(x)

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