QUESTION IMAGE
Question
24 in the diagram below of right triangle abc, altitude
\overline{bd} is drawn.
which ratio is always equivalent to \cos a?
- \frac{ab}{bc}
- \frac{bd}{bc}
- \frac{bd}{ab}
- \frac{bc}{ac}
Step1: Recall the cosine definition
In a right - triangle, \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\). In right - triangle \(ABC\) with right angle at \(B\), \(\cos A=\frac{AB}{AC}\). Also, since \(\triangle ABC\sim\triangle ADB\) (by AA similarity: \(\angle A=\angle A\) and \(\angle ABC = \angle ADB=90^{\circ}\)).
Step2: Use similarity of triangles
For similar triangles \(\triangle ABC\) and \(\triangle ADB\), the ratios of corresponding sides are equal. In right - triangle \(ADB\) with right angle at \(D\), \(\cos A=\frac{AD}{AB}\). Also, in right - triangle \(ABC\) and right - triangle \(BDC\), \(\angle A+\angle C = 90^{\circ}\) and \(\angle DBC+\angle C=90^{\circ}\), so \(\angle A=\angle DBC\).
In right - triangle \(ABC\), \(\cos A=\frac{AB}{AC}\). In right - triangle \(BDC\), \(\cos\angle DBC=\frac{BD}{BC}\). Since \(\angle A=\angle DBC\), \(\cos A=\frac{BD}{BC}\)
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- \(\frac{BD}{BC}\)