Discover Circle Theorem Help for Triangle PST: PSR Angle = 77 Degrees

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In triangle PST, angle PSR is calculated as 180 degrees minus the sum of the other two angles, which are 81 and 22 degrees, resulting in PSR being 77 degrees. The discussion confirms that this calculation is correct based on the principle that the sum of angles in a triangle equals 180 degrees. The context involves a circle with points PQRS and straight lines PQT and SRT. The approach to finding angle PSR adheres to the relevant geometric equations. The conclusion affirms the accuracy of the calculation.
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Homework Statement
Find angle PSR, if PQRS is a circle and PQT and SRT are straight lines
Relevant Equations
sum of angles in a triangle = 180 degree
IMG_20230610_014211 (1).jpg

From triangle PST, angle PSR = 180-(81+22)
=77 degrees sum of angles in triangle.
Is my attempt correct?
 
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Chijioke said:
Homework Statement: Find angle PSR, if PQRS is a circle and PQT and SRT are straight lines
Relevant Equations: sum of angles in a triangle = 180 degree

View attachment 329703
From triangle PST, angle PSR = 180-(81+22)
=77 degrees sum of angles in triangle.
Is my attempt correct?
Looks good to me.
 
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Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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