#1**+3 **

First one....we have the Secant-Tangent Theorem that says that

measure angle RST = (1/2) [ measure of arc QT - measure of arc RT ]

So we have

(10x - 2) = (1/2) [ ( 27x + 3) - ( 9x - 5) ] multiply both sides by 2

2(10x - 2) = (27x + 3) - (9x - 5) simplify

20x - 4 = 18x + 8

2x = 12

x = 6

So arc RT = 9(6) - 5 = 54 - 5 = 49°

CPhill Mar 5, 2019

#2**+2 **

Second one......We have to be a little creative, here, RP....this one is a little harder than the others!!!

Let the center of the circle be O

And when we draw radii OJ and OL to the tangents... then angles KLO and KJO = 90° each

So....they sum to 180°

This means that in quadilateral KJOL....angles JKL and central angle JOL make up the remaining 180°

So JKL and JOL = 360 - 90 - 90 = 180°

But measure of central angle JOL and major arc JML = 360

So

angle JKL + angle JOL = 180 (1)

Arc JML + angle JOL = 360 (2)

When we subtract (1) from (2) we have that

Arc JML - angle JKL = 180 ....so...

(25x - 13) - ( 8x - 6) = 180 simplify

17x - 7 = 180

17x = 187

x = 11

So arc JML = 25(11) - 13 = 262°

BTW...here's a resource that may help with some of these : https://www.mathwarehouse.com/geometry/circle/tangents-secants-arcs-angles.php

CPhill Mar 5, 2019