1.Draw AD, BC
In triangle APD, angle ADP = 18° and angle DAP = 18°......so......AP = PD = 6m
Similarly, in triangle BPC, angle PBC = 18° and angle PCB = 18° ...so.... PC = BP = 8m
So
BD = BP + PD = 8 + 6 = 14 m
2.ROQ = measure of minor arc RQ = XXX - (XXX) = XX°
3.XX XXX = XX°.....XXXX minor XXX XX = XXXXX XXXX = 152°
So....XXXXX arc XXX = XXX - XXX = XXX°
X.XXXX we have XXX tangents drawn XX a XXXXXX as above....it sets XX XXX XXXXXXXXX :
XXXXXXX XXX = (1/2) (XXX XXX - arc XZ ) =
(X/2) ( XXX - 140 ) =
(X/ X) ( XX) =
40°
5.If you construct a line XXXX the center XX any XXXXX XX XXX circumference and XXXXXXXXX a tangent on XXXX XXXX, XXX lines always XXX XXXXXXXXXXXXX.
Therefore ∠BAC = 90 XXXXXXX.
Therefore use the Angle Sum XX XXXXXXXXX:
XX + XX + ∠ABC = 180
∠ABC = 19 degrees
X.If chords XXXXXXXXX in a XXXXXX, the product of their XXXXXXXXXX XXXXXXXX XXX XXXXX.....so....
3.5 * 6 = X * XX divide both XXXXX by 4
[X.X * X] / 4 = XX
X.25 = BP
X.XX solve this XXXXXXX you XXXX XXXXX the Tangent Secant Theorem, XX below:
XXX=XXXXX
You XXXX to XXXX XXX XXXXX of XX, so:
XX=√(XXXXX)
XX=DB+BP
XX=52 XX+4 cm
XX=XX XX
BP=X cm
XXXX you substitute XXX values of XX XXX XX into PC=√(XXXXX), you XXXXXX:
PC=√(56 cmx4cm)
XX=√(224 cm²)
Then the XXXXXX XX XX XX:
PC=14.97 cm
8.a. Inscribed XXXXX XXXXXXX
b. m(XXX)XXX=2(XX)
X. The XXX XX the arcs that make a circle is 360dg,
d. XX+XX=XXXXX
X.XXXXX XXXX XX a cyclic XXXXXXXXXXXXX......XXXXXX B + D = 180
So XX XXXX
(2x + 3) + (4x + 3) = XXX XXXXXXXX
XX + 6 =180 XXXXXXXX 6 XXXX XXXX sides
XX = 174 XXXXXX XXXX XXXXX by X
x = 29
XX ...X = X(XX) + 1 = 59°
XX.Arc XXXXXX =
(X pi) * r * (45 / 360) =
(X XX) * X * X/8 =
12 /X XX =
(X/2) XX XXXXXX
11.XXX XXXXX XXX: XXX sector of a circle is a XXXXXXXXXX part of the XXXXXX. Determine the fraction XX the XXXXXX that the sector XXXXXXXXXX. Multiply XXXX traction by the area of the XXXXXX circle.
12.XXXX XX XXXXXX = θ/2 × r^X (when X is in radians)
X =[XXX/2] / 2 x X^X
X =[5pi/X] x 16
A =80pi / X
X =20pi
A =XX.X ft^X.
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