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anemone
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If $Q$ is a point on the altitude $AM$ of triangle $ABC$, and that $\angle QBA=20^{\circ}$, $\angle QBC=40^{\circ}$ and $\angle QCB=30^{\circ}$, find $\angle QCA$.
anemone said:If $Q$ is a point on the altitude $AM$ of triangle $ABC$, and that $\angle QBA=20^{\circ}$, $\angle QBC=40^{\circ}$ and $\angle QCB=30^{\circ}$, find $\angle QCA$.
Albert said:Point Q must be the orthocenter of
$\triangle ABC $
we have :$\angle QCA+30+40=90$
$\therefore \angle QCA=20^o$
The altitude of a triangle is a line segment drawn from a vertex perpendicular to the opposite side, or to the line containing the opposite side. It forms a right angle with the side it intersects.
To find the altitude of a triangle, you can use the formula "altitude = (2 * area) / base". This formula applies to all types of triangles, whether they are equilateral, isosceles, or scalene.
The altitude of a triangle divides the triangle into two right triangles. The angle opposite the altitude is the complementary angle to the angle at the vertex where the altitude is drawn. This means that the sum of these two angles is always 90 degrees.
To find the angle QCA, you can use the trigonometric functions sine, cosine, or tangent, depending on the given information. For example, if you know the length of the altitude and the adjacent side, you can use the cosine function to find the angle QCA.
Yes, you can still find the angle QCA even if the altitude AM is outside of the triangle. You can use the Pythagorean theorem to find the length of the adjacent side, and then use the appropriate trigonometric function to find the angle QCA.