Math / Trigonometry

QuestionTrigonometry triangles ID: 20369739
In triangle $A B C$ , the measure of angle $B$ is $90^{\circ}$ and $\overline{B D}$ is an altitude of the triangle. The length of $\overline{A B}$ is 15 and the length of $\overline{A C}$ is 23 greater than the length of $\overline{A B}$ . What is the value of $\frac{B C}{B D}$ ? A. $\frac{15}{38}$ B. $\frac{15}{23}$ PDF de-sign PDF forms \& eements C. $\frac{23}{15}$ D. $\frac{38}{15}$ trial 5:15 PM Search 11/27/2024

Studdy Solution

STEP 1

1. Triangle $ABC$ is a right triangle with $\angle B = 90^\circ$ .
2. $\overline{BD}$ is the altitude from $B$ to $\overline{AC}$ .
3. The length of $\overline{AB}$ is $15$ .
4. The length of $\overline{AC}$ is $23$ greater than the length of $\overline{AB}$ .

STEP 2

1. Determine the length of $\overline{AC}$ .
2. Use the Pythagorean theorem to find the length of $\overline{BC}$ .
3. Use the geometric mean theorem to find the length of $\overline{BD}$ .
4. Calculate $\frac{BC}{BD}$ .

STEP 3

Determine the length of $\overline{AC}$ :
Since $\overline{AC}$ is $23$ greater than $\overline{AB}$ , we have:
$AC = AB + 23 = 15 + 23 = 38$

STEP 4

Use the Pythagorean theorem to find $\overline{BC}$ :
In $\triangle ABC$ , since $\angle B = 90^\circ$ , we have:
$AC^2 = AB^2 + BC^2$
Substitute the known values:
$38^2 = 15^2 + BC^2$
$1444 = 225 + BC^2$
$BC^2 = 1444 - 225$
$BC^2 = 1219$
$BC = \sqrt{1219}$

STEP 5

Use the geometric mean theorem to find $\overline{BD}$ :
The geometric mean theorem states that in a right triangle, the altitude to the hypotenuse is the geometric mean of the segments it divides the hypotenuse into. Let $AD = x$ and $DC = y$ . Then:
$BD^2 = AD \cdot DC$
Since $AD + DC = AC = 38$ , and using the properties of the right triangle, we have:
$BD = \frac{AB \cdot BC}{AC}$
Substitute the known values:
$BD = \frac{15 \cdot \sqrt{1219}}{38}$

STEP 6

Calculate $\frac{BC}{BD}$ :
$\frac{BC}{BD} = \frac{\sqrt{1219}}{\frac{15 \cdot \sqrt{1219}}{38}}$
Simplify the expression:
$\frac{BC}{BD} = \frac{38}{15}$
The value of $\frac{BC}{BD}$ is:
$\boxed{\frac{38}{15}}$

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