+0

# plz help asap

0
464
2
+2448

Mar 5, 2019

#1
+106532
+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°

Mar 5, 2019
#2
+106532
+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

Mar 5, 2019