in the adjoining figure, PT and PT' are tangents from P to the circle with centre O. If angle TOT'=30° , then find angle OPT
Answers
Answer:
SOLUTION :
\begin{gathered} \sf \frac{y - 1}{4} - \frac{y + 2}{3} = 4 \\ \end{gathered}
4
y−1
−
3
y+2
=4
\begin{gathered} \sf \: \sf \implies \: 12 \times ( \frac{y - 1}{4} - \frac{y + 2}{3} )= 4 \times 12 \\ \end{gathered}
⟹12×(
4
y−1
−
3
y+2
)=4×12
\sf \implies 3y - 3 - 4y - 8 = 48⟹3y−3−4y−8=48
\sf \implies \: y = - 59⟹y=−59
Answer:
25°
Step-by-step explanation:
Angle T = 90°
Angle T' = 90° - Radius Perpendicular to tangent.
Angle TOT' + Angle OTP + Angle OT'P + Angle TPT' = 360° - Angle sum property of quadrilateral
130° + 90° + 90° + Angle TPT' = 360
310° + Angle TPT' = 360°
Angle TPT' = 50° - (1)
P bisects OP so,
Angle OPT = Angle OPT' - (2)
Angle OPT + Angle OPT' = 50° - From (1)
2 Angle OPT = 50° - From (2)
Angle OPT= 50/2
Angle OPT = 25°.