Exercise 48 Page 820

We are given that

m ∠ E = 6 2^{\circ}

and

d = 38 .

We are asked to find a value for the side length of

e

such that no triangle

D E F

exists. To do so, we will consider the values which are less than the length of the height of the triangle and greater than zero, because it is a side length and must be non-negative.

0 < e < h

Let's imagine a figure that satisfies this condition.

We will first find the length of

h

and then make the possible values of

e

less than

h .

Having an acute angle

6 2^{\circ}

and length of the hypotenuse in a right triangle, we can find the opposite side length

h

by using the sine ratio.

sin θ = \frac{Opposite Side}{Hypotenuse}

Now, we will substitute

θ = 6 2^{\circ}

and

hypotenuse = 38

into the formula to find the

opposite

side

h .

sin θ = \frac{Opposite Side}{Hypotenuse}

SubstituteValues

Substitute values

sin 62 = \frac{h}{3 8}

▼

Solve for

h

MultEqn

$LHS \cdot 38 = RHS \cdot 38$

38 \cdot sin 62 = h

RearrangeEqn

Rearrange equation

h = 38 \cdot sin 62

UseCalc

Use a calculator

h = 33.552008 \dots

RoundDec

Round to $2$ decimal place(s)

h \approx 33.55

Great! Then, we can write an inequality to express the possible

e

values which give no solution for

D E F .

0 < e < h \Rightarrow 0 < e < 33.55

Finally, we can arbitrarily choose

e = 30

as our solution.