We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website.
Please click on "Accept cookies" if you agree to the setting of cookies. Cookies that do not require consent remain unaffected by this, see
cookie policy and privacy policy.
DECLINE COOKIES

#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