Line RS ( 150 cms ) is a common base to 2 triangles: RSQ and RSP.
Point P is located south of Q.
Ang PRQ = 43 , Ang QRP = 37 \(\displaystyle \color{blue}{\angle PRQ\text{ and }\angle QRP\text{ are the same angle!}}\)
Ang PSR = 50.
Find the length of PQ, Ang RQS, and length PR.
Should we start by dropping a vert line from P to RS
and solve for Ang P in the small triangle ?
FIRST, congratulations; you actually have thought about your problem and make a suggestion on how to approach it. That is what we like to see.
As soroban has said, angle prq is the same as angle qrp so you have implied that 43 = 37, which is nonsense.
And what is angle p? Please correct those errors.
There is also an ambiguity in your paraphrase of the problem. Point p is located "south of q." Do you mean that p is located within triangle rsq? Do you mean that q is above line rs and p below line rs? In either case, do you mean that p and q are located on the same perpendicular to rs? Please clarify your meaning. I recognize that you may be looking at a problem with a diagram, but we cannot see your diagram so you must describe the problem even more carefully than your text.
Line \(\displaystyle RS\) (150 cm) is a common base to 2 triangles, \(\displaystyle RSQ\) and \(\displaystyle RSP.\)
Point \(\displaystyle P\) is located neat the center of \(\displaystyle \Delta RSQ.\) .\(\displaystyle \angle PSQ = 43^o, \angle RSP = 50^o, \angle QRP = 37^o.\)
Find \(\displaystyle \angle RQS\) and the lengths of \(\displaystyle PQ\) and \(\displaystyle PR.\)
The area of \(\displaystyle \Delta RSP\) looks to be half the area of \(\displaystyle \Delta RQS.\) .Does this help? . Not really.
The base split would be 75. .No, \(\displaystyle \Delta PRS\) is not isosceles.
Now use the sine rule?
Angle R is split. .What does this mean?
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Q
o
* *
* *
* *
P
* o *
37[SUP]o[/SUP]* *
* * **
*43[SUP]o[/SUP] 50[SUP]o [/SUP]*
R o * * * * * o S
150[/SIZE][/SIZE]