In ΔPQR, the side QR is extended to S such that RS = PR. If ∠QPS = 110° and ∠PRQ = 70°, then the value of ∠PQR is:

In ΔPQR, the side QR is extended to S such that RS = PR. If ∠QPS =

1.	In ΔPQR, the side QR is extended to S such that RS = PR. If ∠QPS = 110° and ∠PRQ = 70°, then the value of ∠PQR is:
A.	45°
B.	40°
C.	50°
D.	35°
Answer» E.

Discussion

No Comment Found

Related MCQs

PQ and RS are common tangents to two circles intersecting at A and B. AB, when produced both sides, meet the tangents PQ and RS at X and Y, respectively. If AB = 3 cm, XY = 5 cm, then PQ (in cm) will be
If the sum of all interior angles of a regular polygon is twice the sum of all its exterior angles then the polygon is
In an isosceles right-angled triangle, the perimeter is 30 m. Find its area (Approximate)
In ΔPQR, S and T are mid-points of PQ and PR, respectively. If ∠QPR = 75° and ∠PRQ = 40°, then ∠TSQ is:
In the circle below, chord AB is extended to meet the tangent DC at D. If AB = 12 cm and DC = 8 cm, find the length of BD.
If the base and height of a triangle is doubled and tripled respectively. Find the difference between the final and initial area of the triangle.
In a circle with centre O, AC and BD are two chords. AC and BD meet at E when produced. If AB is the diameter and ∠AEB = 68°, then the measure of ∠DOC is:
In the following figure (not to scale), AB = CD and \(\rm\overline{AB}\) and \(\rm\overline{CD}\) are produced to meet at point P. If ∠BAD = 70°, then find ∠P
Consider the following statements:1. The diagonals of a trapezium divide each other proportionally.2. Any line drawn parallel to the parallel sides of a trapezium divides the non-parallel sides proportionally.Which of the above statements is / are correct?
DEF is an isosceles triangle with such that DE = DF = 60 cm and EF = 96 cm. DG is a median to base EF. What is the length (in cm) of DG?

In ΔPQR, the side QR is extended to S such that RS = PR. If ∠QPS = 110° and ∠PRQ = 70°, then the value of ∠PQR is:

Discussion

No Comment Found

Related MCQs

Reply to Comment