In the given figure, PT and PS are two tangents drawn
from an external point P to a circle with centre O and
radius r. If OP = 2r, show that angle OTS = angle OST = 30°.
(please don't use trigonometry to find Angle TOQ and angle SOQ)(CLASS 10)
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3
Answer:
Given: OT=OS=r and OP=2r
In ΔTOP,
sinTPO=
OP
TO
=
2r
r
=
2
1
Since, sin30
o
=
2
1
Therefore, ∠TPO=30
o
Similarly for ∠OPS=30
o
Now,
∠TPS=∠TPO+∠OPS
= 30
o
+30
o
=60
o
As we know that ∠TPS+∠TOS=180
o
So, ∠TOS=180
o
−∠TPS
= 180
o
−60
o
=120
o
Now, in ΔTOS, let ∠OST=∠OTS=x
o
Also, ∠TOS+x
o
+x
o
=180
o
120
o
+2x
o
=180
o
2x
o
=60
o
x
o
=30
o
Therefore, ∠OST=∠OTS=30
o
.
Answered by
0
Answer:
In Fig, from an external point P, two tangents PT and PS are drawn to a circle with centre O and radius r. If OP=2r, show that ∠ OTS = ∠ OST = 30°. ∴ ΔOTP is a 30o-60o-90o, right triangle.
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