In the figure above, TP is a Tangent to the circle TRQ with centre O. If <TPO=28⁰ and <ORQ=15⁰, find;
1) <PQT
2) <QTO
Unfortunately, since I am unable to see the figure, I cannot provide specific measurements or angles. However, I can provide general formulas and concepts that can be used to find the values you are looking for.
1) To find <PQT, you can use the fact that the tangent to a circle is perpendicular to the radius at the point of tangency. This means that <TPO and <PQT are complementary angles. So, if <TPO is 28 degrees, then <PQT will be 90 - 28 = 62 degrees.
2) To find <QTO, you can use the fact that the angle formed by two tangents drawn from an external point to a circle is equal to half the difference of the intercepted arcs. In this case, the intercepted arc by angle ORQ is the arc QR. So, if <ORQ is 15 degrees, then <QTO will be half the difference between the degrees of the intercepted arc and a full circle (360 degrees). If you know the measure of the intercepted arc QR, you can subtract it from 360 and divide the result by 2 to find <QTO.
To find the measures of angles <PQT and <QTO, we can use the properties of tangents and the angles in a circle.
1) <PQT:
Since TP is tangent to the circle, <TPO is a right angle (90⁰). We know that <TPO = 28⁰, so the remaining angle <OPT can be found by subtracting the given angle from 90⁰:
<TPO = 28⁰
<TPO = 90⁰ - <OPT
28⁰ = 90⁰ - <OPT
Next, we can solve for <OPT by rearranging the equation:
<OPT = 90⁰ - 28⁰
<OPT = 62⁰
Since <POT is an exterior angle to triangle OPQ, we can find <PQT by subtracting <OPT from <QOP:
<QOP = <POT - <OPT
<QOP = 90⁰ - 62⁰
<QOP = 28⁰
Therefore, <PQT = <QOP = 28⁰.
2) <QTO:
We know that <TPO = 28⁰ and <ORQ = 15⁰. Since angles on a straight line add up to 180⁰, <OPT can be found by subtracting the given angles from 180⁰:
<TPO + <ORQ + <OPT = 180⁰
Substituting the given values:
28⁰ + 15⁰ + <OPT = 180⁰
Now we can solve for <OPT by rearranging the equation:
<OPT = 180⁰ - 28⁰ - 15⁰
<OPT = 137⁰
Since <QOT is an exterior angle to triangle OQT, we can find <QTO by subtracting <OPT from <QOT:
<QOT = <QTO + <OPT
<QOT = 180⁰ - 137⁰
<QOT = 43⁰
Therefore, <QTO = <QOT = 43⁰.
To find the angles, we can start by using the properties of tangents.
1) To find angle <PQT, we need to use the fact that tangents are perpendicular to the radius drawn to the point of tangency.
Since TP is a tangent to the circle at point P, angle <TPO is 90 degrees. We also know that angle <TPO is 28 degrees, so we can set up the following equation:
90 degrees = 28 degrees + <PQT
Solving for <PQT:
<PQT = 90 degrees - 28 degrees
<PQT = 62 degrees
Therefore, <PQT is 62 degrees.
2) To find angle <QTO, we can use the fact that the sum of the angles around a point is 360 degrees.
The angle <ORQ is 15 degrees. We know that angle <QTO is supplementary to angle <ORQ, so they add up to 180 degrees. Using this information, we can set up the following equation:
180 degrees = 15 degrees + <QTO
Solving for <QTO:
<QTO = 180 degrees - 15 degrees
<QTO = 165 degrees
Therefore, <QTO is 165 degrees.