Math, asked by ishikagoel246, 4 months ago

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​

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Answered by Anonymous
0

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

Answered by rushikhalate6
1

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°.

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