Chapter 4.8, Problem 16E

To determine

Tofind:the altitude of the isosceles triangle shown in the figure and round off to two decimal.

Answer to Problem 16E

The altitude of an isosceles triangle is a=1.6 meters.

Explanation of Solution

Given:

${\text{θ=18}}^0$ and $b=10$

Concept used:

Isosceles triangle has two equal sides and angle.

  $\text{cosθ=}\frac{\text{Base}}{\text{Hypotenuse}}$

Pythagoras theorem:

  ${\left(\text{Perpindicular}\right)}^2+{\left(\text{Base}\right)}^{\text{2}}={\left(\text{Hypotenuse}\right)}^2$

Calculation:

Consider the value ${\text{θ=18}}^0$ and $b=10$ meters for the isosceles triangle.

To find the altitude of an isosceles triangle, draw a perpendicular on BC $\left(\text{let b}\right)$ ,so that b is divided into two halves let it be x.

Let the altitude be a and hypotenuse be c.

To solve for c, use the cosine formula in triangle ABD.

  $\text{cosB=}\frac{\text{Base}}{\text{Hypotenuse}}$

  $\text{cosB=}\frac{\text{x}}{\text{c}}$

  ${\text{cos18}}^0\text{=}\frac{5}{\text{c}}$

  $c=\frac{5}{\cos {18}^0}$

Now put the value of cosine

  $c=\frac{5}{0.9510}$

To find the altitude of an isosceles triangle use the Pythagoras theorem;

  ${\left(\text{Perpindicular}\right)}^2+{\left(\text{Base}\right)}^{\text{2}}={\left(\text{Hypotenuse}\right)}^2$

Here

  ${\text{a}}^{\text{2}}{\text{+x}}^{\text{2}}{\text{=c}}^{\text{2}}$

  ${\text{a}}^{\text{2}}{\text{+5}}^{\text{2}}\text{=}{\left(5.26\right)}^{\text{2}}$

  ${\text{a}}^{\text{2}}\text{+25=27}\text{.67}$

Now, take square root

  $\text{a=}\sqrt{27.67-25}$

  $a=1.6$

Hence, the altitude of an isosceles triangle is a=1.6 meters.