external tangent

[logistic_guy]

\(\displaystyle \bold{Given}\): \(\displaystyle \text{\large$\odot$}\text{A}\) is tangent to \(\displaystyle \text{\large$\odot$}\text{B}\) at \(\displaystyle \text{R}\). \(\displaystyle \overline{\text{PT}}\) is a common external tangent at \(\displaystyle \text{P}\) and \(\displaystyle \text{T}\). \(\displaystyle \angle\text{Q} = 43^{\circ}\). \(\displaystyle \bold{Find}\) \(\displaystyle \angle\text{S}\).

 

[khansaheb]

\(\displaystyle \bold{Given}\): \(\displaystyle \text{\large$\odot$}\text{A}\) is tangent to \(\displaystyle \text{\large$\odot$}\text{B}\) at \(\displaystyle \text{R}\). \(\displaystyle \overline{\text{PT}}\) is a common external tangent at \(\displaystyle \text{P}\) and \(\displaystyle \text{T}\). \(\displaystyle \angle\text{Q} = 43^{\circ}\). \(\displaystyle \bold{Find}\) \(\displaystyle \angle\text{S}\).
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Please share your work/thoughts about this problem

 

[logistic_guy]

\(\displaystyle m\angle Q = \frac{1}{2}m\overset{\large\frown}{PR}\)

\(\displaystyle 43^{\circ} = \frac{1}{2}m\overset{\large\frown}{PR}\)

\(\displaystyle m\overset{\large\frown}{PR} = 2 \times 43^{\circ} = 86^{\circ}\)

 

[Aion]

\(\displaystyle m\angle Q = \frac{1}{2}m\overset{\large\frown}{PR}\)

\(\displaystyle 43^{\circ} = \frac{1}{2}m\overset{\large\frown}{PR}\)

\(\displaystyle m\overset{\large\frown}{PR} = 2 \times 43^{\circ} = 86^{\circ}\)

Here’s my approach: Let [imath]\angle Q=\alpha[/imath] and [imath]\angle S =\beta[/imath]. By the central angle theorem, we have [imath]\angle PAR=2\alpha[/imath] and [imath]\angle TBR=2\beta[/imath]. Since [imath]AP \perp PT[/imath] and [imath]BT \perp PT[/imath] it follows that [imath]AP \parallel BT[/imath]. Then, by the parallel postulate, [math]2\alpha +2\beta=180\degree \implies \beta=90\degree-\alpha=47\degree[/math]

 

[logistic_guy]

Here’s my approach: Let [imath]\angle Q=\alpha[/imath] and [imath]\angle S =\beta[/imath]. By the central angle theorem, we have [imath]\angle PAR=2\alpha[/imath] and [imath]\angle TBR=2\beta[/imath]. Since [imath]AP \perp PT[/imath] and [imath]BT \perp PT[/imath] it follows that [imath]AP \parallel BT[/imath]. Then, by the parallel postulate, [math]2\alpha +2\beta=180\degree \implies \beta=90\degree-\alpha=47\degree[/math]

Your way is a little bit different than what I am thinking to do. My idea is based on something that I have learnt from @jonah2.0 as we discussed a similar sketch a long time ago. Watch and learn Aion.

 

[logistic_guy]

\(\displaystyle m\angle PAR = m\overset{\large\frown}{PR} = 86^{\circ}\)

 

[logistic_guy]

If I create a point on the segment \(\displaystyle \overline{AP}\), say \(\displaystyle C\), such that the segment \(\displaystyle \overline{BC} \parallel \overline{PT}\), then \(\displaystyle \Delta ABC\) is a right triangle with \(\displaystyle m\angle ABC = 90^{\circ} - 86^{\circ} = 4^{\circ}\).

 

[Aion]

If I create a point on the segment \(\displaystyle \overline{AP}\), say \(\displaystyle C\), such that the segment \(\displaystyle \overline{BC} \parallel \overline{PT}\), then \(\displaystyle \Delta ABC\) is a right triangle with \(\displaystyle m\angle ABC = 90^{\circ} - 86^{\circ} = 4^{\circ}\).

Sure, you’re almost done! Notice that [imath]\angle TBR = \angle ABC + \angle CBT = 4^{\circ} + 90^{\circ} = 94^{\circ}[/imath], so [imath]\angle S = ?[/imath]

 

[logistic_guy]

Sure, you’re almost done!

I am an honest person. It was not my idea to draw the segment \(\displaystyle \overline{BC}\). It was stolen from the great @jonah2.0

 

[Agent Smith]

[imath]0^{\circ} < S < 90^{\circ}[/imath]

 

[logistic_guy]

The segment \(\displaystyle \overline{BC}\) passes the circumference of circle \(\displaystyle \text{\large$\odot$}\text{B}\) say at point \(\displaystyle D\), then

\(\displaystyle m\overset{\large\frown}{RD} = m\Delta RBD = m\Delta ABC = 4^{\circ}\)

 

[khansaheb]

The segment \(\displaystyle \overline{BC}\) passes the circumference of circle \(\displaystyle \text{\large$\odot$}\text{B}\) say at point \(\displaystyle D\), then

\(\displaystyle m\overset{\large\frown}{RD} = m\Delta RBD = m\Delta ABC = 4^{\circ}\)

Where is "C"?

 

[logistic_guy]

Where is "C"?

\(\displaystyle m\angle CBT = m\overset{\large\frown}{CT} = 90^{\circ}\) since \(\displaystyle \overline{CB} \parallel \overline{PT}\).

 

[Agent Smith]

A change request: Please don't post gemoetry questions or post them with the relevant diagram. I WAS very good at geometry although I remember losing 5 whole marks because I couldn't solve a circle-based problem.

I believe in TIMTOWTDI

 

[logistic_guy]

A change request: Please don't post gemoetry questions or post them with the relevant diagram. I WAS very good at geometry although I remember losing 5 whole marks because I couldn't solve a circle-based problem.

I believe in TIMTOWTDI

If I stop posting Geometry, Aion will be sad. But my goal is to make every student happy. Therefore, I can't comply with your request.

 

[logistic_guy]

\(\displaystyle m\angle CBT = m\overset{\large\frown}{CT} = 90^{\circ}\) since \(\displaystyle \overline{CB} \parallel \overline{PT}\).

\(\displaystyle \textcolor{red}{\bold{Correction.}}\)
\(\displaystyle m\angle CBT = m\overset{\large\frown}{\textcolor{green}{D}T} = 90^{\circ}\) since \(\displaystyle \overline{CB} \parallel \overline{PT}\).

Or

\(\displaystyle m\angle DBT = m\overset{\large\frown}{DT} = 90^{\circ}\) since \(\displaystyle \overline{DB} \parallel \overline{PT}\).

 

[logistic_guy]

And here is the final magical touches.

\(\displaystyle \bold{Find}\) \(\displaystyle \angle\text{S}\).

\(\displaystyle \angle\text{S} = \frac{1}{2}m\overset{\large\frown}{RT} = \frac{1}{2}\left(m\overset{\large\frown}{RD} + m\overset{\large\frown}{DT}\right) = \frac{1}{2}\bigg(4^{\circ} + 90^{\circ}\bigg) = \frac{1}{2}\bigg(94^\circ\bigg) = 47^{\circ}\)