PB is a chord of the circle and POC is its diameter and PT is the tangent, such that 2PCB = 35º. If PT is the tangent to the circle at the point P, then find the measure of BPT.
Answers
Answer:
The \angle BAT∠BAT = 50°.
Step-by-step explanation:
We are given that AB is a chord of a circle and AOC is its diameter such that \angle∠ ACB = 50°.
Also, AT is the tangent to the circle at point A.
As we can clearly see in the figure that \triangle△ ABC is a right-angled triangle with right angle at B. So, the sum of all angles of a triangle is equal to 180°.
In \triangle△ ABC ;
\angle∠ ABC + \angle∠ BAC + \angle∠ ACB = 180°
90° + \angle∠ BAC + 50° = 180°
140° + \angle∠ BAC = 180°
\angle∠ BAC = 180° - 140° = 40°.
Now, we are given that AT is the tangent to the circle at point A, this means;
\angle∠ BAC + \angle∠ BAT = 90° {beacuse tangent is perpendicular to the radius}
40° + \angle∠ BAT = 90°
\angle∠ BAT = 90° - 40° = 50°.
Given, PQ is a chord of a circle with center O.
Also, ∠QPT=60°.
Let x be the point on the tangent PT.
∠QPT+∠OPT=90
⇒∠OPT=30
0
−∠QPT=90
0
−60
0
=30
0
In ΔPOQ
∠POQ=180−(∠OPQ+∠PQO)=180−30−30=120
0
Minor arc ∠POQ=120
0
Therefore Major arc ∠POQ=360
0
−120
0
=240
0
Angle subtended by an arc at centre is double the angle subtended by it on remaining part of circle
∴∠QRP=
2
1
∠POQ=120
0