As the title says, I'm having trouble with tangent perpendicular lines and circles in triangles. I was given the problem linked above for practice & given the answer, but I am still stumped on it.
I was only able to solve for angle a, which equals 80, since 60 + 40 + a = 180.
The given answers to the equation are:
- Angle a = 80
- Angle b = 40
- Angle c = 60
I'm more confused on how to solve for angle b & c in the first place. I would also greatly appreciate a breakdown of the steps of solving this problem if possible. Thanks in advance.
$\endgroup$ 23 Answers
$\begingroup$the angles in the tangent, they are incribed angles too, the same as c and b
Improving what i said, the angle c = the angle 60 and the angle 40 = to the angle b, because the angle formed by the tangent and a segment passing through the circuference twice it is an inscribed angle.
$\endgroup$ 1 $\begingroup$Illustration just for one angle, $\beta$:
\begin{align} \angle EAO&=\angle OFA=90^\circ ,\\ \angle ABC&=\tfrac12\angle AOC= \angle AOF=\angle FOC=\beta ,\\ \angle FAO&=90^\circ-\beta ,\\ \angle EAF&=\angle EAO-\angle FAO =90^\circ-(90^\circ-\beta)= \beta . \end{align}
The reasoning for the other angle ($\gamma)$ is similar.
$\endgroup$ $\begingroup$One of Euclid's propositions..
In a circumscribed triangle the angle between a chord and tangent at a point equals the angle in the alternate segment.
That determines where the angles $ 40^0, 60^0 $ should go.
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