Chapter 5.4, Problem 21E

(a)

To determine

To find: The length of a leg of the triangle.

Answer to Problem 21E

  $\text{The length of leg = 5}\text{.4 m}$

Explanation of Solution

Given information: Geometry: Each base angle of an isosceles triangle measures ${42}^{\circ }30\text{'}$ . The base is 14.6 meters long.

Calculation:

The triangle we will work with (the right half)

To find the length of the leg we are working with adjacent and hypotenuse so we use cosine.

  $\begin{array}{l}\cos {42.5}^{\circ }=\frac{7.3}{x}\\ x=\frac{7.3}{\cos {42.5}^{\circ }}\\ x=5.4\text{ m}\end{array}$

(b)

To determine

To find: The altitude of the triangle.

Answer to Problem 21E

  $\text{Altitude = 6}\text{.7 m}$

Explanation of Solution

Given information: Geometry: Each base angle of an isosceles triangle measures ${42}^{\circ }30\text{'}$ . The base is 14.6 meters long.

Calculation:

This is the right half of the triangle with the altitude making the left side.

The leg length has been solved as 5.4 in part (a).

We are looking for the altitude. Using our angle we can set this up as either a tangent or sine problem. I will use tangent as I prefer to use as many given numbers as I can.

That way, if I’ve made one mistake (say on the leg length) it will not cause me to make more. So the equation is,

  $\begin{array}{l}\tan {42.5}^{\circ }=\frac{a}{7.3}\\ a=7.3\times \tan {42.5}^{\circ }\\ a=6.7\end{array}$

Thus the altitude is 6.7 m.

(c)

To determine

To find: the area of the triangle.

Answer to Problem 21E

  $\text{Area = 48}{\text{.8 m}}^2$

Explanation of Solution

Given information: Geometry: Each base angle of an isosceles triangle measures ${42}^{\circ }30\text{'}$ . The base is 14.6 meters long.

Calculation:

This is the original triangle. We found the altitude to be 6.69 in part (b) of this problem. The base is given as 14.6. The formula for the area of a triangle is,

  $\begin{array}{l}A=\frac{bh}{2}\\ A=\frac{14.6\times 6.69}{2}=48.8{\text{ m}}^2\end{array}$

Thus the area of the triangle is $48.8{\text{ m}}^2$