A circle is drawn inside a triangle ABC to touch the sides BC,CA and AB at P,Q and R respectively.if A=56° and B=68°, find the angles of triangle PQR
answer
To find the angles of triangle PQR, we need to use some geometric properties of circles and triangles.
First, let's label the points where the circle touches the sides of the triangle: P on BC, Q on CA, and R on AB.
We can observe that the line PR is a diameter of the circle. This is because the center of the circle lies on the perpendicular bisector of PR (since it is equidistant from P and R), and any line passing through the center of a circle and terminating on the circle is a diameter.
Since PR is a diameter, the angle PQR is a right angle (90 degrees). This is a property of the diameter of a circle.
Now, let's focus on the angles at points P, Q, and R.
Since the circle is drawn inside the triangle, we know that the tangent line at each point of contact is perpendicular to the radius of the circle passing through that point. Therefore, the angles formed by these perpendiculars are right angles.
Angle APQ = 90 - Angle QAP
Similarly, angle BPQ = 90 - Angle QBP
And angle CRQ = 90 - Angle QCR
To find the angles, we need to determine the measures of Angle QAP, Angle QBP, and Angle QCR.
Using the given information A = 56° and B = 68°, we can find Angle QAP and Angle QBP.
Angle QAP = Angle B - Angle BAP = Angle B - 90 - A = 68 - 90 - 56 = -18° (note that the negative angle arises because it lies below the horizontal line)
Angle QBP = Angle A - Angle ABP = 56 - 90 - (180 - B) = 56 - 90 - 180 + 68 = -146°
Since angles in a triangle sum up to 180 degrees, we can calculate Angle QCR as:
Angle QCR = 180 - Angle QAP - Angle QBP = 180 - (-18) - (-146) = 180 + 18 + 146 = 344°
Therefore, the angles of triangle PQR are:
Angle PQR = 90°
Angle QRP = -18°
Angle RPQ = -146°