Question

GIVEN:
PB tangent
PV, PU secants

If m VU= 80° and m ST= 40°, then ∠1 =
If m VU= 70° and m ST= 30°, then ∠2 =
If m VB= 60° and m BS = 30°, then ∠3 =
If VS = 9, SP = 12 and UT = 4, then TP =

Answer 1

Part a) If m VU= 80° and m ST= 40°, then ∠1 =

we know that

The measurement of the external angle is the semi-difference of the arcs it comprises.

so

$Angle (1)=\frac{1}{2}*(80\°-40\°)=20\°$

therefore

the answer Part a) is

∠1 =$20\°$

Part b) If m VU= 70° and m ST= 30°, then ∠2 =

we know that

The measure of the internal angle is the semi- sum of the arcs comprising it and its opposite.

so

$Angle (2)=\frac{1}{2}*(70\°+30\°)=50\°$

the answer Part b) is

∠2 =$50\°$

Part c) If m VB= 60° and m BS = 30°, then ∠3 =

we know that

The measurement of the external angle is the semi-difference of the arcs it comprises.

so

$Angle (3)=\frac{1}{2}*(60\°-30\°)=15\°$

therefore

the answer Part c) is

∠3 =$15\°$

Part d) If VS = 9, SP = 12 and UT = 4, then TP =

we know that

The Intersecting Secants Theorem  states that when two secant lines intersect each other outside a circle, the products of their segments are equal.

so

$PT*PU=PS*PV$

we have

$PS=12\ units\\PV=PS+VS=12+9=21\ units\\PU=PT+UT=PT+4$

substitute

$PT*(PT+4)=12*21$

$PT^{2} +4PT-252=0$

using a graphing tool------> to resolve the second order equation

see the attached figure

the solution is

$PT=14\ units$

therefore

the answer Part d) is

$PT=14\ units$

Answer 2

Answer:

1= 40

2=50

3=15

TP= 14