Question

help please. friends
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Answer 1

❏ Question:-

@ for details please refer to the questionar attachment,

Find $\angle PTQ=?$ if $\angle POQ=110°$

❏ Solution:-

Given:-

• $\angle POQ=110\degree$

To Find:-

• $\angle PTQ=?$

Now, we know that tangent to a point of a circle is $\perp$ to the radius at that point.

$\therefore PO\perp PT\: \:and \:\:OQ\perp QT$

$\therefore \angle OPT=\angle OQT=90\degree$

Now, from the quadrilateral POQT,

$\sf\bf\implies \angle OPT+\angle POQ+\angle OQT +\angle PTQ =360\degree$

$\sf\bf\implies 90\degree+110\degree+90\degree +\angle PTQ =360\degree$

$\sf\bf\implies 290\degree +\angle PTQ =360\degree$

$\sf\bf\implies \angle PTQ =360\degree-290\degree$

$\sf\bf\implies\boxed{\large{ \red{\angle PTQ =70\degree}}}$ (ans).

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$\underline{ \huge\mathfrak{hope \: this \: helps \: you}}$

Answer 2

❏ Question:-

Refer to attachment

❏ Solution:-

Given:-

• $\angle POQ=110\degree$

To Find:-

• $\angle PTQ=?$

Now, we know that tangent to a point of a circle is $\perp$ to the radius at that point.

$\therefore PO\perp PT\: \:and \:\:OQ\perp QT$

$\therefore \angle OPT=\angle OQT=90\degree$

Now, from the quadrilateral POQT,

$\sf\bf\implies \angle OPT+\angle POQ+\angle OQT +\angle PTQ =360\degree$

$\sf\bf\implies 90\degree+110\degree+90\degree +\angle PTQ =360\degree$

$\sf\bf\implies 290\degree +\angle PTQ =360\degree$

$\sf\bf\implies \angle PTQ =360\degree-290\degree$

$\sf\bf\implies\boxed{\large{ \red{\angle PTQ =70\degree}}}$ (ans).

━━━━━━━━━━━━━━━━━━━━━━━

$\underline{ \huge\mathfrak{hope \: this \: helps \: you}}$