Question

PQR is a tangent to circle QABCD. AB || QD. CB=CD. Let 2, -30° and D₂ =70°. P A 1.1 Calculate Q₁. 1.2 Prove that C=110 A 1.3 Calculate B1​

Answer 1

To solve the problem, we need to use the properties of tangents to circles and the fact that AB || QD.

1.1. Calculate Q₁:

Since PQR is tangent to the circle at point Q, we know that angle PQR is 90 degrees. We also know that angle QRD is 70 degrees (given in the problem). Since AB || QD, we know that angle QAB is equal to angle QRD (alternate angles). Therefore, angle QAB is also 70 degrees.

Using the fact that the angles in a triangle add up to 180 degrees, we can calculate angle AQB as follows:

angle AQB = 180 - angle QAB - angle QBA
= 180 - 70 - 90
= 20 degrees

Since angle QAB is 70 degrees, we can use the fact that the angles in a triangle add up to 180 degrees to find angle QBC:

angle QBC = 180 - angle BQC - angle QCB
= 180 - 90 - angle QCB
= 90 - angle QCB

Since CB = CD, we know that angle QBC = angle QCD. Therefore, we can write:

angle QBC = angle QCD = (180 - angle D₂)/2 = (180 - 70)/2 = 55 degrees

Substituting this into the equation we derived earlier for angle QBC, we get:

55 = 90 - angle QCB

Solving for angle QCB, we get:

angle QCB = 35 degrees

Finally, we can use the fact that the angles in a triangle add up to 180 degrees to find angle BQA:

angle BQA = 180 - angle AQB - angle QBA
= 180 - 20 - 90
= 70 degrees

Therefore, Q₁ = angle BQA - angle QCB = 70 - 35 = 35 degrees.

1.2. Prove that C=110:

Since AB || QD, we know that angle AQB is equal to angle QCD (alternate angles). Therefore, we can write:

angle QCD = angle AQB = 20 degrees

We also know that CB = CD. Therefore, triangle BCD is an isosceles triangle, and we can write:

angle BDC = angle CBD = (180 - angle D₂)/2 = (180 - 70)/2 = 55 degrees

Using the fact that the angles in a triangle add up to 180 degrees, we can write:

angle BCD = 180 - angle CBD - angle BDC
= 180 - 55 - 55
= 70 degrees

Therefore, C = 180 - angle QCD - angle BCD = 180 - 20 - 70 = 90 degrees.

Since CB = CD, we know that angle CBD is also equal to (180 - angle D₂)/2 = 55 degrees. Therefore, we can write:

angle BDA = 180 - angle AQB - angle CBD
= 180 - 20 - 55
= 105 degrees

Therefore, C + BDA = 90 + 105 = 195 degrees. Since CB = CD, we know that angle CBD is also equal to angle BDC = 55 degrees. Therefore, we can write:

angle BDC = 180 - angle BDA - angle ADB
= 180 - 105 - angle ADB

Solving for angle ADB, we get:

angle ADB = 20 degrees

Therefore, C = angle ADB + angle CBD = 20 + 55 = 75 degrees