MathWorks 10 Workbook Solutions Chapter 7 Trigonometry of Right Triangles 134

3. Consider ∆ABC. The right angle is ∠B. Next, rearrange the equation to solve for y.

a2 + c2 = (x + y)2 x 2 + y2 = z 2

Consider ∆ABD. The right angle is ∠D. y2 = z 2 − x 2

x2 + z2 = c2
y= z2 − x2

Consider ∆BDC. The right angle is ∠D. 6. First consider the bottom triangle, with legs of
4.2 cm and 6.8 cm and hypotenuse x.
z2 + y2 = a2
c2 = a 2 + b 2
4.
x 2 = 4.22 + 6.82
x 2 = 17.64 + 46.24
x 2 = 63.88
ℓ h x = 63.88
x ≈ 8.0 cm

Next, consider the upper triangle, with legs x

d
and y, and a hypotenuse of 10.4 cm.

The distance from the base of the house, d, and a 2 + b 2 = c2

the side of the house, h, form the legs of the y2 + x 2 = 10.4 2
triangle. The ladder, ℓ, forms the hypotenuse. y2 + ( 63.88)2 = 108.16
The relationship between the sides can be
y2 + 63.88 = 108.16
expressed as follows.
y2 = 108.16 − 63.88
h2 + d2 = ℓ2
y2 = 44.28
5. Rearrange the equation to solve for x.
y= 44.28
x 2 + y2 = z 2 y ≈ 6.7 cm
x 2 = z 2 − y2

x = z 2 − y2

Note: Generally when you take the square root

of a number, there are two possible solutions
(e.g., x2 = 4, x = ±2). In this case, since x is
a length, you do not need to consider the
negative solution.
MathWorks 10 Workbook Solutions Chapter 7 Trigonometry of Right Triangles 135

7. Let the length of the ladder be ℓ, the height of PRACTISE YOUR NEW SKILLS, p. 295
the ladder be h, and distance from the base of
the house be d. 1. Convert the dimensions of the stairway from
feet and inches to inches.
h 2 + d 2 = 2
6′4 ′′ = [(6 × 12) + 4]′′
382 + d 2 = 402
6′4 ′′ = (72 + 4 )′′
d 2 = 402 − 382
6′4 ′′ = 7 6′′
d = 402 − 382
8′6′′ = [(8 × 12) + 6]′′
d = 1600 − 1444
8′6′′ = (96 + 6 )′′
d = 156
8′6′′ = 10 2′′
d ≈ 12.5 ft
Let the diagonal length of the stairway be s, the
The foot of the ladder is approximately 12.5 feet rise be r, and the horizontal distance be h.
from the base of the house.
s2 = r 2 + h 2
8. Calculate the diagonal distance across the field.
s 2 = 762 + 1022
2 + w 2 = d 2
s= 762 + 1022
180 + 120 = d
2 2 2

s= 5776 + 10 404
180 + 120 = d
2 2

s = 16 180
32 400 − 14 400 = d
s ≈ 127.2 in
46 800 = d
The diagonal length of the stairway is about
216.3 ≈ d 127.2 inches.
If you walked diagonally across the field, your Convert to feet and inches.
route would be about 216.3 m long.
127.2 ÷ 12 = 10.6 ft
Calculate the distance you would travel if you 10.6 ft = 10 ft (0.6 × 12) in
walked around the edge of the field.
10.6 ft = 10 ft 7.2 in
dedge =  + w
The diagonal length of the stairway is about
dedge = 180 + 120
10 ft 7.2 in.
dedge = 300 m

Calculate the difference in route lengths.

300 − 216.3 = 83.7 m

Your route would be about 83.7 m shorter by

walking diagonally across the field.
MathWorks 10 Workbook Solutions Chapter 7 Trigonometry of Right Triangles 136

A LT ERN AT I V E SOL U T ION 2. Let the length of the guy wire be g, the height
of the tower be t, and the distance from the
Convert the measurements to feet.
base of the tower be d.

( )
6′4′′ = 6 4
12
′
d2 + t2 = g 2

6′4′′ ≈ 6.3′ d 2 + 24 2 = 282

d 2 = 282 − 24 2
( )
8′6′′ = 8 6
12
′
d = 282 − 24 2
8′6′′ = 8.5′
d = 784 − 576
s2 = r 2 + h 2 d = 208
s 2 = 6.32 + 8.52 d ≈ 14.4 m

s= 6.32 + 8.52 The guy wire is attached to the ground about

s= 39.69 + 72.25 14.4 m from the base of the tower.

s = 111.94 3. Let the rise of the ramp be h, the run be r, and

s ≈ 10.6 ft the length of the ramp be ℓ.

The diagonal length of the stairway is 2 = h 2 + r 2

about 10.6 ft. 2 = 3.52 + 10.52

Convert to feet and inches. = 3.52 + 10.52

127.2 ÷ 12 = 10.6 ft  = 12.25 + 110.25

10.6 ft = 10 ft (0.6 × 12) in  = 122.5
10.6 ft = 10 ft 7.2 in   ≈ 11.1 m

The diagonal length of the stairway is about The ramp is approximately 11.1 m long.
10 ft 7.2 in.
MathWorks 10 Workbook Solutions Chapter 7 Trigonometry of Right Triangles 137

4. Let the height of the TV be h, the width be w, 5. Calculate how far north the boat sailed.
and the diagonal distance between opposite
3 h × 12 km/h = 36 km
corners be d.
Calculate how far east the boat sailed.
h2 + w2 = d2
2 h × 18 km/h = 36 km
322 + w 2 = 522
w 2 = 522 − 322 Let n be the distance travelled north, e be the
distance travelled east, and d be the distance
w = 522 − 322
from the starting point.
w = 2704 − 1024
d2 = n 2 + e2
w = 1680
d 2 = 362 + 362
w ≈ 41.0 in
d = 362 + 362
The TV is about 41 inches wide.
d = 1296 + 1296
d = 2592
d ≈ 50.9 km

The boat is about 50.9 km from its starting

point.