Download MPM 2D Solving Right Triangles

MPM2D Grade 10 Academic
SOLVING RIGHT TRIANGLES
Practice
1. Solve each triangle. Round each side length to the nearest tenth of a unit,
and each angle to the nearest degree.
a)
Solution
∠C = 90o – ∠E = 90o – 33o = 57o,
𝐶𝐷
𝐶𝐷
sin ∠E = ⟹ sin 33o =
𝐶𝐸
22
CD = 22 sin 33o ≈12.0 m,
𝐸𝐷
𝑅𝐷
cos ∠E = ⟹ cos 33o =
𝐶𝐸
22
ED = 22 cos 33o ≈18.5 m.
b)
Solution
∠B = 90o – ∠G = 90o – 65o = 25o,
𝐵𝐹
29
sin ∠G = ⟹ sin 65o =
𝐵𝐺
BG =
tan ∠G =
𝐵𝐹
𝐺𝐹
29
𝐵𝐺
≈ 32.0 cm,
sin 65°
29
o
⟹ tan 65 =
GF =
𝐺𝐹
29
tan 65°
≈ 13.5 cm.
c)
Solution
∠P = 90o – ∠R = 90o – 29o = 61o,
𝑅𝑀
51
cos ∠R =
⟹ cos 29o =
𝑅𝑃
RP =
tan ∠R =
𝑃𝑀
𝑅𝑀
51
𝑅𝑃
≈ 58.3 m,
cos 29°
𝑃𝑀
o
⟹ tan 29 =
51
PM = 51 tan 29o ≈ 28.3 m.
d)
Solution
∠X = 90o – ∠Z = 90o – 30o = 60o,
𝑋𝑌
6
sin ∠Z = ⟹ sin 30o =
𝑋𝑍
XZ =
tan ∠Z =
𝑋𝑌
𝑌𝑍
6
sin 30°
6
o
⟹ tan 30 =
YZ =
𝑋𝑍
6
≈ 12 cm,
𝑌𝑍
tan 30°
≈ 10.4 cm.
e)
Solution
∠S = 90o – ∠N = 90o – 54o = 36o,
𝑂𝑆
𝑂𝑆
sin ∠N = ⟹ sin 54o =
𝑁𝑆
17
OS = 17 sin 54o ≈13.8 km,
𝑂𝑁
𝑂𝑁
cos ∠N =
⟹ cos 54o =
𝑁𝑆
17
ON = 17 cos 54o ≈10.0 km.
f)
Solution
∠T = 90o – ∠M = 90o – 13o = 77o,
𝑇𝑋
𝑇𝑋
sin ∠M =
⟹ sin 13o =
𝑇𝑀
13
TX = 13 sin 13o ≈ 2.9 m,
𝑋𝑀
𝑋𝑀
cos ∠M =
⟹ cos 13o =
𝑇𝑀
13
XM = 13 cos 13o ≈ 12.7 m,
2. Find all the unknown angles, to the nearest degree, and all the unknown
side lengths, to the nearest tenth of a unit.
a)
Solution
𝐴𝐵
13
cos ∠A =
⟹ cos ∠A =
𝐴𝐶
–1
13
15
∠A = cos (15) ≈ 30 ,
∠C = 90o – ∠A ≈ 90o – 30o = 60o.
o
By the Pythagorean Theorem, determine BC:
BC = √𝐴𝐶 2 − 𝐴𝐵2 = √152 − 132 = √56 ≈ 7.5 cm.
b)
Solution
tan ∠G =
𝐼𝐻
𝐺𝐻
⟹ tan ∠G =
9
9
7
∠G = tan –1 (7) ≈ 52o,
∠I = 90o – ∠G ≈ 90o – 52o = 38o.
By the Pythagorean Theorem, determine GI:
GI = √𝐼𝐻 2 + 𝐺𝐻 2 = √92 + 72 = √130 ≈ 11.4 cm.
c)
Solution
𝐾𝐿
5
cos ∠L =
⟹ cos ∠L =
𝐽𝐿
6
5
∠L = cos –1 (6) ≈ 34o,
∠J = 90o – ∠L ≈ 90o – 34o = 56o.
By the Pythagorean Theorem, determine JK:
JK = √𝐽𝐿2 − 𝐾𝐿2 = √62 − 52 = √11 ≈ 3.3 cm.
d)
Solution
𝑄𝑅
tan ∠P =
⟹ tan ∠P = 1.1
𝑄𝑃
–1
∠P = tan (1.1) ≈ 48o,
∠R = 90o – ∠P ≈ 90o – 48o = 42o.
By the Pythagorean Theorem, determine PR:
PR = √𝑄𝑅 2 + 𝑄𝑃2 = √112 + 102 = √221 ≈ 14.9 cm.
e)
Solution
𝑇𝑆
tan ∠R =
⟹ tan ∠R = 1.125
𝑅𝑆
∠R = tan –1 (1.125) ≈ 48o,
∠T = 90o – ∠R ≈ 90o – 48o = 42o.
By the Pythagorean Theorem, determine TR:
TR = √𝑇𝑆 2 + 𝑅𝑆 2 = √182 + 162 = √580 ≈ 24.1 m.
f)
Solution
sin ∠W =
𝑋𝑌
𝑊𝑌
⟹ sin ∠W =
25
25
54
∠W = sin –1 (54) ≈ 28o,
∠Y = 90o – ∠W ≈ 90o – 28o = 62o.
By the Pythagorean Theorem, determine WX:
WX = √𝑊𝑌 2 − 𝑋𝑌 2 = √542 − 252 = √56 ≈ 47.9 cm.
Application and Problem Solving
3. Peggy’s Cove Lighthouse, in Nova Scotia, is possibly the most
photographed lighthouse in the world. The observation deck is about 20 m
above sea level. From the observation deck, the angle of depression of a boat
is 60 . How far is the boat from the lighthouse, to the nearest metre?
Solution
𝐵𝐶
20
tan A = ⟹ tan 6o =
AB =
𝐴𝐵
20
tan 6°
𝐴𝐵
≈ 190 m.
The boat is about 190 m
from the lighthouse.
4. A kite is 32 m above the ground. The angle the kite string makes with the
ground is 390 . How long is the kite string, to the nearest metre?
Solution
𝐴𝐵
32
sin ∠C = ⟹ sin 39o =
AC =
𝐴𝐶
32
sin 39°
𝐴𝐶
≈ 51 m.
The kite string ia about 51 m long.
5. The world’s fastest roller coaster is at Six Flags Magic Mountain. At the
start, riders are shot forward and then up a tower. They then begin the
backward descent. From a point 100 m from the foot of the tower, the angle
of elevation of the top of the tower is 520 . Find the height of the tower, to the
nearest metre.
Solution
ℎ
tan 52o =
⟹ h = 100 tan 52o ≈ 128 m
100
The height of the tower is about 128 m.
6. Montreal’s Marathon building is 195 m tall. From a point level with, and
48 m from, the base of the building, what is the angle of elevation of the top
of the building, to the nearest degree?
Solution
195
195
tan x =
⟹ x = tan –1 ( ) ≈ 79o
38
38
The angle of elevation of the top of the building
is about 79o.
7. Ropes are used to pull a totem pole upright. Then, the ropes are anchored
in the ground to hold the pole until the hole is filled. One of the ropes holding
this totem pole is 18 m long and forms an angle of 480 with the ground.
Find, to the nearest metre,
a) the height of the totem pole
Solution
ℎ
sin 48o = ⟹ h = 18 sin 48o ≈ 13 m
18
The height of the totem pole is about 13 m.
b) how far the anchor point is from the
base of the totem pole
Solution
𝑥
cos 48o = ⟹ x = 18 cos 48o ≈ 12 m.
18
The anchor point is about 12 m from the base of the totem pole.
8. The two guy wires supporting a flagpole are each anchored 7 m from the
flagpole and form an angle of 520 with the ground. What is the total length
of guy wire, to the nearest metre, needed to support this flagpole?
Solution
7
7
cos 52o = ⟹ x =
≈ 11.4 m.
𝑥
cos 52°
l = 2x = 2(11.4) ≈ 23 m.
The total length of guy wire is about 23 m.
9. Edmonton’s CN Tower is a high-rise office building. It has 27 storeys.
From a point 35 m from the base of the building and level with the base, the
angle of elevation of the top is 72.5o.
a) Find the height of Edmonton’s CN Tower, to the nearest metre.
Solution
ℎ
tan 72.5o = ⟹ h = 35 tan 72.5o≈ 111 m.
35
The height of Edmonton’s CN Tower is about 111 m.
b) Toronto’s CN Tower is a tourist attraction, with a height
of 555 m. If an office building were the height of Toronto’s
CN Tower, how many stories would you expect it to have?
Explain.
Solution
𝑛
27
=
⟹ n = 27× 5 = 135 m
555 111
I would expect it has 135 stories.
10. A coast guard patrol boat is 14.8 km east of the Brier Island lighthouse. A
disabled yacht is 7.5 km south of the lighthouse.
a) How far is the patrol boat from the yacht, to the nearest tenth of a
kilometre?
Solution
By Pythagorean Theorem, determine BY:
BY2 = 7.52 + 14.82
BY2 = 275.29
BY ≈16.6 km
The patrol boat from is 16.6 km from the yacht.
b) At what angle south of due west, to the nearest degree, should the patrol
boat travel to reach the yacht?
Solution
7.5
tan 𝜃 =
≈ 0.506757
14.8
𝜃 = tan –1 (0.506757) ≈ 27o
To reach the yacht, the patrol boat should travel in the direction W27o S.
11. A sign shows that a hill has a grade of 9%. What angle does the hill make
with the horizontal, to the nearest tenth of a degree?
Solution
tan 𝜃 = 0.09 ⟹ 𝜃 = tan –1 (0.09) ≈ 5.1o.
12. Use right triangles ABC and DEF to complete the table. Leave all ratios
in fraction form.
Triangle
ABC
DEF
tan x sin x cos x tan (90° – x)
5
5
12
/12
/13 12/13
/5
3
3
4
4
/4
/5
/5
/3
a) How is tan x related to tan (90° – x)?
Answer
tan x × tan (90° – x) = 1
b) How is sin x related to cos (90° – x)?
Answer
sin x = cos (90° – x)
c) How is cos x related to sin (90° – x)?
Answer
cos x = sin (90° – x)
sin (90° – x) cos (90° – x)
12
5
/13
/13
4
3
/5
/5