18 x 1 /2 x -3 x -2/3
1 answer:
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Step-by-step explanation:
Hey there!
By looking at the figure, the given "2x" and "6x + 28" are co-interior angle.
So,
2x + 6x + 28 = 180°. ( sum of co-interior angle is 180°)
8x + 28°=180°
8x = 180° - 28°
8x = 152°
X = 19°
<u>There</u><u>fore</u><u>,</u><u> </u><u>X </u><u>=</u><u> </u><u>1</u><u>9</u><u>°</u><u>.</u>
<em><u>Hope</u></em><em><u> </u></em><em><u>it</u></em><em><u> helps</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>
Answer:
Part 1) m∠1 =(1/2)[arc SP+arc QR]
Part 2)
Part 3) PQ=PR
Part 4) m∠QPT=(1/2)[arc QT-arc QS]
Step-by-step explanation:
Part 1)
we know that
The measure of the inner angle is the semi-sum of the arcs comprising it and its opposite.
we have
m∠1 -----> is the inner angle
The arcs that comprise it and its opposite are arc SP and arc QR
so
m∠1 =(1/2)[arc SP+arc QR]
Part 2)
we know that
The <u>Intersecting Secant-Tangent Theorem,</u> states that the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment.
so
In this problem we have that
Part 3)
we know that
The <u>Tangent-Tangent Theorem</u> states that if from one external point, two tangents are drawn to a circle then they have equal tangent segments
so
In this problem
PQ=PR
Part 4)
we know that
The measurement of the outer angle is the semi-difference of the arcs it encompasses.
In this problem
m∠QPT -----> is the outer angle
The arcs that it encompasses are arc QT and arc QS
therefore
m∠QPT=(1/2)[arc QT-arc QS]