A chord PQ and a diameter RS of a circle (with centre O) are produced to intersect each other at a point T and QT is equal to the radius of the circle. If angle PTR = 42°, then the measure of angle POR is:
To determine the measure of angle POR, we will use properties of circles, triangles, and the theorem related to angles formed by two secants intersecting outside a circle. Let 'r' denote the radius of the circle with center O.
Understanding the Given Geometry
We are given a circle with center O. PQ is a chord, and RS is a diameter. The lines containing chord PQ and diameter RS are produced (extended) to intersect at a point T outside the circle. We are also given two crucial pieces of information:
- QT is equal to the radius of the circle, i.e., \(QT = r\).
- The angle \(\angle PTR = 42^\circ\).
From the definition of a circle, all radii are equal. Thus, \(OP = OQ = OR = OS = r\).
Analyzing Triangle OQT
Consider the triangle \(\triangle OQT\):
- We know that \(OQ = r\) (since OQ is a radius).
- We are given that \(QT = r\).
- Since \(OQ = QT = r\), \(\triangle OQT\) is an isosceles triangle.
- In an isosceles triangle, the angles opposite the equal sides are equal. Therefore, the angle opposite OQ (\(\angle OTQ\)) is equal to the angle opposite QT (\(\angle QOT\)).
- The angle \(\angle PTR\) is given as \(42^\circ\). Since P, Q, T are collinear and R, S, T are collinear, \(\angle OTQ\) is the same as \(\angle PTR\).
- So, \(\angle OTQ = 42^\circ\).
- From the isosceles property of \(\triangle OQT\), we have \(\angle QOT = \angle OTQ = 42^\circ\).
Relating Central Angle QOS to QOT
The diameter is RS, and O is the center. This means R, O, S are collinear points. The problem states that the diameter RS is "produced" to intersect at T. This implies that T lies on the line containing the diameter RS. Given the common arrangement in such problems and the provided options, the most likely configuration for the collinear points T, S, O, R is T-S-O-R. In this arrangement, the ray OT and the ray OS are in the same direction.
- Since T, S, O, R are collinear in the order T-S-O-R, the ray OS is along the same direction as the ray OT.
- Therefore, the central angle \(\angle QOS\) (formed by OQ and OS) is equal to \(\angle QOT\) (formed by OQ and OT).
- From our previous calculation, we found \(\angle QOT = 42^\circ\).
- Thus, \(\angle QOS = 42^\circ\).
Applying Secant Theorem for External Intersection
When two secant lines (or produced chords/diameters) intersect outside a circle at a point T, the measure of the angle formed at T is half the difference of the measures of the intercepted arcs. In this case, the line PQT is a secant intersecting the circle at P and Q, and the line RST is a secant intersecting the circle at R and S.
- The angle at the external point T is \(\angle PTR\).
- The intercepted arcs are arc PR and arc QS.
- The measures of these arcs correspond to their respective central angles: \(\text{arc PR} = \angle POR\) and \(\text{arc QS} = \angle QOS\).
- According to the theorem: \[\angle PTR = \frac{1}{2} (\text{measure of arc PR} - \text{measure of arc QS})\] \[42^\circ = \frac{1}{2} (\angle POR - \angle QOS)\]
Calculating Angle POR
Now, substitute the known value of \(\angle QOS = 42^\circ\) into the equation:
\[42^\circ = \frac{1}{2} (\angle POR - 42^\circ)\]Multiply both sides by 2:
\[2 \times 42^\circ = \angle POR - 42^\circ\] \[84^\circ = \angle POR - 42^\circ\]Add \(42^\circ\) to both sides to solve for \(\angle POR\):
\[\angle POR = 84^\circ + 42^\circ\] \[\angle POR = 126^\circ\]Therefore, the measure of angle POR is \(126^\circ\).
Important Questions from Miscellaneous
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A railway wagon (open at the top) of mass M1 is moving with speed v1 along a straight track. As a result of rain, after some time it gets partially filled with water so that the mass of the wagon becomes M2 and speed becomes v2. Taking the rain to be falling vertically and the water stationery inside the wagon, the relation between the two speeds v1 and v2 is :
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1. Distance between the longitudes becomes zero on North Pole and South Pole.
2. Distance between the longitudes is maximum on the Equator.
3. Number of longitudes is more than number of latitudes.
Which of the statements given above is/are correct?
One block of 2⋅0 kg mass is placed on top of another block of 3⋅0 kg mass. The coefficient of static friction between the two blocks is 0⋅2. The bottom block is pulled with a horizontal force F such that both the blocks move together without slipping. Taking acceleration due to gravity as 10 m/s2, the maximum value of the frictional force is :
