In the figure, the length of the chord AB, if PA=6 cm and angle PAB-60 degrees is
Answers
Answer:
Consider a circle C with center O.
We have PA and PB are tangents of the circle, PA=10cm and
∠APB=60
o
Join OP,
Such that,
In △PAC and △PBC we have,
PA=PB [tangent of the circle fro the outer point p is equal]
∠PAC=∠PBC [angle made by the external tangent on a
circle is equal]
PC=CP [common]
so,
△PAC ≅ △PBC [By SAS criteria]
so,
AC=BC ........(i)
∠ACP=∠BCP ......(ii)
since
∠APB=∠APC+∠BPC
so,
∠APC=
2
1
×60
0
=30
0
∠APC=30
0
∠ACP+∠BCP=180
0
from equation 2 we get
∠ACP=
2
1
×180
0
=90
0
Thus in Right △ ACP
sin30
0
=
AP
AC
2
1
=
10
AC
cm
AC=5cm
Since
AC=BC
so,
AB=AC+BC
=5cm+5cm
=10cm
solution
Answered By
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Answer:
Consider a circle C with center O.
We have PA and PB are tangents of the circle, PA=10cm and
∠APB=60
o
Join OP,
Such that,
In △PAC and △PBC we have,
PA=PB [tangent of the circle from the outer point p is equal]
∠PAC=∠PBC [angle made by the external tangent on a
circle is equal]
PC=CP [common]
so,
△PAC ≅ △PBC [By SAS criteria]
so,
AC=BC .( i )
∠ACP=∠BCP .(ii)
since
∠APB=∠APC+∠BPC
so,
∠APC=
2
1
×60
0
=30
0
∠APC=30
0
∠ACP+∠BCP=180
0
from equation 2 we get
∠ACP=
2
1
×180
0
=90
0
Thus in Right △ ACP
sin30
0
=
AP
AC
2
1
=
10
AC
cm
AC=5cm
Since
AC=BC
so,
AB=AC+BC
=5cm+5cm
=10cm