Chapter 5.4, Problem 8CFU

(a)

To determine

To find: The altitude of the triangle.

Answer to Problem 8CFU

  $\text{Altitude = 8}\text{.24}$

Explanation of Solution

Given information: Geometry: Each base angle of an isosceles triangle measures ${55}^{\circ }30\text{'}$ . Each of the congruent sides is 10 centimeters long.

Calculation:

To find the altitude I have taken the right side of the left triangle in the sketch and redrawn it to the right.

We are looking for the opposite side and know the hypotenuse so this is a sine problem.

I also change the $30\text{'}\text{ to 1/2}$ a degree and added to the ${55}^{\circ }$ . So our angle is ${55.5}^{\circ }$ .

  $\begin{array}{l}\sin {55.5}^{\circ }=\frac{\text{Altitude}}{10}\\ \sin {55.5}^{\circ }\times 10=\text{Altitude}\\ \text{Altitude}=0.8241\times 10\\ \text{Altitude}=8.24\end{array}$

(b)

To determine

What is the length of the base?

Answer to Problem 8CFU

  $\text{Base = 11}\text{.3}$

Explanation of Solution

Given information: Geometry: Each base angle of an isosceles triangle measures ${55}^{\circ }30\text{'}$ . Each of the congruent sides is 10 centimeters long.

Calculation:

To find the base I have taken the right side of the left triangle in the sketch and redrawn it to the right.

We are looking for the adjacent side and know the hypotenuse so this is a cosine problem.

I also change the $30\text{'}\text{ to 1/2}$ a degree and added to the ${55}^{\circ }$ . So our angle is ${55.5}^{\circ }$ . Once we find the small side of the triangle on the right we need to double it to find the base of the original triangle.

  $\begin{array}{l}\cos {55.5}^{\circ }=\frac{\text{Adjacent}}{10}\\ \cos {55.5}^{\circ }\times 10=\text{Adjacent }\left(\text{small side}\right)\\ \text{Adjacent = }0.5664\times 10=5.66\end{array}$

Now double this to get the base of the triangle

  $\begin{array}{l}\text{Base = 2}\times \text{5}\text{.66}\\ \text{Base = 11}\text{.3}\end{array}$

(c)

To determine

To find: The area of the triangle.

Answer to Problem 8CFU

  $\text{Area = 46}\text{.6}$

Explanation of Solution

Given information: Geometry: Each base angle of an isosceles triangle measures ${55}^{\circ }30\text{'}$ . Each of the congruent sides is 10 centimeters long.

Calculation:

Now we have simply $A=\frac{bh}{2}$ where the base was found to be 11.3 and the altitude is 8.24 so,

  $\begin{array}{l}A=\frac{\left(11.3\right)\left(8.24\right)}{2}\\ A=\frac{93.1}{2}\\ \text{Area = 46}\text{.6}\end{array}$