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2 answers:
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Check the picture attached.
the secants of the circle, BA and DC, intersect at point P, as shown.
i)
let m(P)=a
ii)
let m(CDA)=b, then the measure of ar CA is 2b, as "the measure of an inscribed angle is half the measure of the arc it intercepts"
iii)
m(DAB)= m(P)+m(CDA)= a+b, as DAB is an exterior angle of the triangle PAD.
iv)
then the measure of arc DB is 2m(DAB)=2(a+b)=2a+2b
v)
This is always true,
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In our problem, there is a slight difference, we have 1 secant and one tangent, but it is just a special case and nothing changes.
As an exercise, we can repeat the steps of the general proof, by first joining U and T.
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by the theorem we just proved,
m(R)=(m(TU)-m(SU))/2
30°=(100°-m(SU))/2
60°=100°-m(SU)
m(SU)=40°
Check the picture attached.
the secants of the circle, BA and DC, intersect at point P, as shown.
i)
let m(P)=a
ii)
let m(CDA)=b, then the measure of ar CA is 2b, as "the measure of an inscribed angle is half the measure of the arc it intercepts"
iii)
m(DAB)= m(P)+m(CDA)= a+b, as DAB is an exterior angle of the triangle PAD.
iv)
then the measure of arc DB is 2m(DAB)=2(a+b)=2a+2b
v)
This is always true,
----------------------------------------------------------------------------------------------------
In our problem, there is a slight difference, we have 1 secant and one tangent, but it is just a special case and nothing changes.
As an exercise, we can repeat the steps of the general proof, by first joining U and T.
----------------------------------------------------------------------------------------------------
by the theorem we just proved,
m(R)=(m(TU)-m(SU))/2
30°=(100°-m(SU))/2
60°=100°-m(SU)
m(SU)=40°