Below is the graph of a trigonometric function. It intersects its midline at (4.7,-3.8) and it has a maximum point at (-4.7,3.8)
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Given: A circle with centre O; PA and PB are two tangents to the circle drawn from an external point P.
To prove: PA = PB
Construction: Join OA, OB, and OP.
It is known that a tangent at any point of a circle is perpendicular to the radius through the point of contact.
OA⊥PA
OB⊥PB
In △OPA and △OPB
∠OPA=∠OPB (Using (1))
OA=OB (Radii of the same circle)
OP=OP (Common side)
Therefor △OPA≅△OPB (RHS congruency criterion)
PA=PB
(Corresponding parts of congruent triangles are equal)
Thus, it is proved that the lengths of the two tangents drawn from an external point to a circle are equal.
The length of tangents drawn from any external point are equal.
So statement is correct
To prove: PA = PB
Construction: Join OA, OB, and OP.
It is known that a tangent at any point of a circle is perpendicular to the radius through the point of contact.
OA⊥PA
OB⊥PB
In △OPA and △OPB
∠OPA=∠OPB (Using (1))
OA=OB (Radii of the same circle)
OP=OP (Common side)
Therefor △OPA≅△OPB (RHS congruency criterion)
PA=PB
(Corresponding parts of congruent triangles are equal)
Thus, it is proved that the lengths of the two tangents drawn from an external point to a circle are equal.
The length of tangents drawn from any external point are equal.
So statement is correct