MHB Angle between the tangents to the circle

[ramz]

View attachment 4819

Hi everyone, I need further explanations about the answer of this problem.
The answer is angle T = 87.9 degrees.

Thanks.

 

[Prove It]

Hi everyone, I need further explanations about the answer of this problem.
The answer is angle T = 87.9 degrees.

Thanks.

Use the cosine rule to evaluate the angle on the origin of the circle.

Radii and tangents to a circle are always perpendicular.

Once you realize this, you have three angles in the quadrilateral OBTA. The angle sum of a quadrilateral is 360 degrees. You should be able to evaluate angle T from there.

 

[ramz]

Ahh. Thank you so much.

 

[ramz]

Ahh. Thank you so much.

 

[MarkFL]

Prove It's method is more succinct, but here's an outline of what I did:

1.) Bisect $\triangle ABO$ and use the Pythagorean theorem to find the altitude.

2.) Use the law of sines to find $\angle OAB$.

3.) Use that fact the a tangent to a circle and a radius to the tangent point are perpendicular to find $\angle BAT$.

4.) Use the fact that $\angle BAT=\angle ABT$ and the sum of interior angles of a triangle being $180^{\circ}$ to find $\angle BTA$.

5.) Use the fact that $\angle BTA+\angle T=180^{\circ}$.

You should find $$\angle T=180^{\circ}\left(1-\frac{2}{\pi}\arccos\left(\frac{\sqrt{301}}{25}\right)\right)\approx87.89^{\circ}$$