Answer:
C) 
Step-by-step explanation:
Use the Angle Addition Postulate to figure this out:

165° = (x + 15)° + (9x)°
165° = (15 + 10x)°
- 15° - 15°
_____________
![\displaystyle \frac{150°}{10} = \frac{[10x]°}{10} \\ \\](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B150%C2%B0%7D%7B10%7D%20%3D%20%5Cfrac%7B%5B10x%5D%C2%B0%7D%7B10%7D%20%5C%5C%20%5C%5C)
[Plug this back into the modeled expression for the
to get the angle measure of 30°]; 
I am joyous to assist you anytime.
Answer:
Part A
The bearing of the point 'R' from 'S' is 225°
Part B
The bearing from R to Q is approximately 293.2°
Step-by-step explanation:
The location of the point 'Q' = 35 km due East of P
The location of the point 'S' = 15 km due West of P
The location of the 'R' = 15 km due south of 'P'
Part A
To work out the distance from 'R' to 'S', we note that the points 'R', 'S', and 'P' form a right triangle, therefore, given that the legs RP and SP are at right angles (point 'S' is due west and point 'R' is due south), we have that the side RS is the hypotenuse side and ∠RPS = 90° and given that
=
, the right triangle ΔRPS is an isosceles right triangle
∴ ∠PRS = ∠PSR = 45°
The bearing of the point 'R' from 'S' measured from the north of 'R' = 180° + 45° = 225°
Part B
∠PRQ = arctan(35/15) ≈ 66.8°
Therefore the bearing from R to Q = 270 + 90 - 66.8 ≈ 293.2°
your answer for this question is y=c hope this helps
RS Bisects QT means, it cuts QT in two equal halves, so QP would be equal to PT, as it is an isosceles triangle, one other side would be equal as well, so RP is equal to PS, Now, by ASA, we conclude the both mentioned triangles in the figure are congruent.
Hope this helps!
Answer:
c not hundred percent sure
Step-by-step explanation: