Chapter 11, Problem 26CUR

To determine

To Find: Ratio in which the altitude to hypotenuse divides the hypotenuse.

Answer to Problem 26CUR

1:3

Explanation of Solution

Given:

A $30°-60°-90°$ triangle with altitude to hypotenuse.

Formula used:

  $\tan \theta =\frac{Perpendicular}{Base}$

Referring to the below figure as per given dimensions:

  

$\Delta CBA$ is a $30°-60°-90°$ triangle with $BD$ as its altitude to the hypotenuse $AC$ .

Let $BD=h,AD=x,DC=y$ .

In $\Delta BDA$ :

  $\begin{array}{l}\tan 60=\frac{BD}{AD}\\ \Rightarrow h=\sqrt{3}x\mathrm{.......}(1)\end{array}$

In $\Delta BDC$ :

  $\begin{array}{l}\tan 30=\frac{BD}{DC}\\ \Rightarrow h=\frac{1}{\sqrt{3}}y\mathrm{.......}(2)\end{array}$

Dividing (1) by (2):

  $x:y=1:3$