OBLIQUE TRIANGLES .pptx

DIVISION TRAINING OF TEACHERS ON LEAST LEARNED
COMPETENCIES IN MATHEMATICS GRADE 9
JANUARY 28- 29, 2021

OBLIQUE TRIANGLE
CARLITO B. DIONEDAS, JR.
Teacher II
Lamba National High School

OBLIQUE TRIANGLES
DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
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Code: 9775547

At the end of the session, teachers should be able to:
• Discuss the nature of oblique triangle;
• Derive the Law of Sines and Cosines
• Solve oblique triangle related problems applying the law of
sines or cosines
OBLIQUE TRIANGLES
OBJECTIVES:

1.Which triangle DOES NOT belong to the group?
OBLIQUE TRAINGLES
a
b
d
c

2. Which of the following is NOT a right triangle?
OBLIQUE TRAINGLES
a
b
d
c

3. Triangle ABC is defined by the following measures:
∠A = 30°, b = 10. What value of side a would lead to
more than one possible triangles?
OBLIQUE TRAINGLES
a. a < 10 c. 10sin30 < a < 10
b. a ≥ 10 sin 30 d. 10 sin 30 ≥ a ≥ 10
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Code: 1254748

4. From the illustration, how high is the building?
OBLIQUE TRAINGLES
a.
b.
d.
c.
ℎ =
sin 95 sin 60
sin 25
𝑖𝑛𝑠𝑢𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑖𝑛𝑓𝑜𝑟𝑚𝑎𝑡𝑖𝑜𝑛
ℎ =
sin 25 sin 60
sin 95
ℎ =
sin 95
sin 25 sin 60
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Code: 7354625

OBLIQUE TRIANGLE
OBLIQUE TRAINGLES
- is a triangle containing NO 90° angle.
Examples:
- Any triangle which is NOT a right triangle.
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Code: 6198069

LAW OF SINES
LAW OF SINES
A
C
B
a
b
c
h

B
a
LAW OF SINES
LAW OF SINES
A
C
b
c
h
𝐬𝐢𝐧 𝑨 =
𝒉
𝒃
𝒃 𝐬𝐢𝐧 𝑨 =
𝒉
𝒃
𝒃 𝐬𝐢𝐧 𝑨 = 𝒉

B
a
LAW OF SINES
LAW OF SINES
A
C
b
c
h
𝐬𝐢𝐧 𝑩 =
𝒉
𝒂
𝒂 𝐬𝐢𝐧 𝑩 =
𝒉
𝒂
𝒂 𝐬𝐢𝐧 𝑩 = 𝒉

LAW OF SINES
LAW OF SINES
𝐬𝐢𝐧 𝑩 =
𝒉
𝒂
𝒉
𝒂
B
a
C
h
c
A
C
B
a
b
c
h

LAW OF SINES
LAW OF SINES
A
C
B
a
b
c
h
𝐬𝐢𝐧 𝑨 =
𝒉
𝒃
𝒃 𝐬𝐢𝐧 𝑨 =
𝒉
𝒃
𝒃 𝐬𝐢𝐧 𝑨 = 𝒉
𝐬𝐢𝐧 𝑩 =
𝒉
𝒂
𝒉
𝒂

LAW OF SINES
LAW OF SINES
A
C
B
a
b
c
h
𝒃 𝐬𝐢𝐧 𝑨 = 𝒉 𝒂 𝐬𝐢𝐧 𝑩 = 𝒉
𝒃 𝐬𝐢𝐧 𝑨 = 𝒉 = 𝒂 𝐬𝐢𝐧 𝑩
𝒃 𝐬𝐢𝐧 𝑨 = 𝒂 𝐬𝐢𝐧 𝑩
𝒂
𝐬𝐢𝐧 𝑨
=
𝒃
𝐬𝐢𝐧 𝑩
=
𝒄
𝐬𝐢𝐧 𝑪
𝒂
𝐬𝐢𝐧 𝑨
=
𝒃
𝐬𝐢𝐧 𝑩
𝑯𝒆𝒏𝒄𝒆,

LAW OF SINES
LAW OF SINES
𝐬𝐢𝐧 𝑨
𝒂
=
𝐬𝐢𝐧 𝑩
𝒃
=
𝐬𝐢𝐧 𝑪
𝒄
Conditions:
1. Given two angles and a side (AAS).
2. Given two sides and an angle (SSA);
𝒂
𝐬𝐢𝐧 𝑨
=
𝒃
𝐬𝐢𝐧 𝑩
=
𝒄
𝐬𝐢𝐧 𝑪
𝒐𝒓

Illustrative Example 1:
LAW OF SINES
25°
P
R
Q p = ?
𝒑
𝒔𝒊𝒏 𝑷
=
𝒒
𝒔𝒊𝒏 𝑸
Solution:
𝒑
𝒔𝒊𝒏 𝟐𝟓
=
𝟐𝟎
𝒔𝒊𝒏 𝟏𝟏𝟓
(𝒔𝒊𝒏 𝟐𝟓)
𝒑
𝒔𝒊𝒏 𝟐𝟓
=
𝟐𝟎
𝒑 =
𝟐𝟎 (𝒔𝒊𝒏 𝟐𝟓)
𝒑 ≈ 𝟗. 𝟑𝟑𝒄𝒎
115°
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Code: 348335

LAW OF SINES
36°
A C
B
?
𝐬𝐢𝐧 𝑪
𝒄
=
𝐬𝐢𝐧 𝑨
𝒂
Solution:
𝐬𝐢𝐧 𝑪
𝟐𝟎
=
𝐬𝐢𝐧 𝟑𝟔
𝟏𝟐
𝟐𝟎
𝐬𝐢𝐧 𝑪
𝟐𝟎
=
𝐬𝐢𝐧 𝟑𝟔
𝟏𝟐
𝐬𝐢𝐧 𝑪 =
𝟐𝟎(𝐬𝐢𝐧 𝟑𝟔)
𝟏𝟐
𝐬𝐢𝐧 𝑪 = 𝟎. 𝟗𝟕𝟗𝟔
𝑪 = 𝒔𝒊𝒏−𝟏(𝟎. 𝟗𝟕𝟗𝟔)
𝑪 ≈ 𝟕𝟖. 𝟒𝟐°
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Code: 188901

Lets Get Real:
LAW OF SINES
Juan and Romella are standing at the seashore
2.5km apart. The coastline is a straight line between
them. Both can see the same ship in the water. The
angle between the coastline and the line between the
ship and Juan is 55 degrees. The angle between the
coastline and the line between the ship and Romella is
65 degrees. How far is the ship from Juan?
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Code: 528717

Illustrating the problem:
LAW OF SINES
2.5 km
Juan
(J)
Romella
(R)
(S)
55° 65°
r=?
Solution:
∠𝑺 = 𝟏𝟖𝟎° − 𝟓𝟓° − 𝟔𝟓°
∠𝑺 = 𝟔𝟎°
𝒓
𝒔𝒊𝒏 𝑹
=
𝒔
𝒔𝒊𝒏 𝑺
𝒓
𝒔𝒊𝒏 𝟔𝟓
=
𝟐. 𝟓
𝒔𝒊𝒏 𝟔𝟎
(𝒔𝒊𝒏 𝟔𝟓)
𝒓
=
𝟐. 𝟓
𝒓 =
𝟐. 𝟓 (𝒔𝒊𝒏 𝟔𝟓)
𝒓 ≈ 𝟐. 𝟔 𝒌𝒎
∴, 𝒕𝒉𝒆 𝒔𝒉𝒊𝒑 𝒊𝒔 𝟐. 𝟔 𝒌𝒎 𝒂𝒘𝒂𝒚 𝒇𝒓𝒐𝒎 𝑱𝒖𝒂𝒏.

Lets Try:
LAW OF SINES
Find the measure of angle B in a triangle ABC given that
∠A = 65° a = 15 cm and b = 20 cm.
𝒔𝒊𝒏 𝑩
𝒃
=
𝒔𝒊𝒏 𝑨
𝒂
𝒔𝒊𝒏 𝑩
𝟐𝟎
=
𝟏𝟓
𝒔𝒊𝒏 𝑩 =
𝟐𝟎(𝒔𝒊𝒏 𝟔𝟓)
𝟏𝟓
𝒔𝒊𝒏 𝑩 =
𝟐𝟎(𝒔𝒊𝒏 𝟔𝟓)
𝟏𝟓
𝒔𝒊𝒏 𝑩 = 𝟏. 𝟐𝟎𝟖𝟒
𝑩 = 𝒔𝒊𝒏−𝟏
(𝟏. 𝟐𝟎𝟖𝟒)
𝑬𝑹𝑹𝑶𝑹
𝑻𝒉𝒆 𝒕𝒓𝒊𝒂𝒏𝒈𝒍𝒆 𝑫𝑶𝑬𝑺 𝑵𝑶𝑻 𝒆𝒙𝒊𝒔𝒕
𝒃𝒂𝒔𝒆𝒅 𝒐𝒏 𝒕𝒉𝒆 𝒈𝒊𝒗𝒆𝒏 𝒄𝒐𝒏𝒅𝒊𝒕𝒊𝒐𝒏
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Code: 8890517

Law of Sines – Ambiguous Case:
LAW OF SINES
The ambiguous case can occur when we are given the angle – side –
side. Consider a triangle in which you are given a, b, and A. (𝒉 = 𝒃 𝒔𝒊𝒏 𝑨)
A is acute…
A
b
a
h
Necessary condition
a < h
Triangles possible
None

LAW OF SINES
A is acute…
A
b a
h
Necessary condition
a = h
Triangles possible
One

LAW OF SINES
A is acute…
A
b a
h
Necessary condition
a ≥ b
Triangles possible
One

LAW OF SINES
A is acute…
A
b a
h
a
Necessary condition
h < a < b
Triangles possible
Two

LAW OF SINES
A is obtuse…
A
b
a
Necessary condition
a ≤ b
Triangles possible
None

LAW OF SINES
A is obtuse…
A
b
a
Necessary condition
a > b
Triangles possible
One

LAW OF SINES
Try this:
A suspension bridge is attached with cables
towards the top of its post in the middle. One of the
cables 72 meters long has an angle of elevation 30
degrees. While the other cable next to it is 62 meters
long. How far from each other are the bases of the
cables (in nearest meter)?
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Code: 3484326

LAW OF SINES
𝟑𝟎°
𝟕𝟐𝒎
𝟔𝟐𝒎
?
Solution:
𝑨 𝑩
𝑪
We need to find m∠ABC in
order to obtain the m∠ACB.
𝑪𝒉𝒆𝒄𝒌 𝒇𝒐𝒓 𝒂𝒎𝒃𝒊𝒈𝒖𝒊𝒕𝒚
𝒉 = 𝟕𝟐(𝒔𝒊𝒏 𝟑𝟎)
𝒉 = 𝟑𝟔𝒎
𝑺𝒊𝒏𝒄𝒆 𝒉 < 𝒂 < 𝒃, 𝒕𝒉𝒆𝒓𝒆 𝒂𝒓𝒆
𝒕𝒘𝒐 𝒑𝒐𝒔𝒔𝒊𝒃𝒍𝒆 𝒕𝒓𝒊𝒂𝒏𝒈𝒍𝒆𝒔
Angle – side – side
𝒉

LAW OF SINES
𝟑𝟎°
𝟕𝟐𝒎
𝟔𝟐𝒎
?
Solution:
𝑨 𝑩
𝑪
𝐬𝐢𝐧 𝑩
𝒃
=
𝐬𝐢𝐧 𝑨
𝒂
𝐬𝐢𝐧 𝑩
𝟕𝟐
=
𝐬𝐢𝐧 𝟑𝟎
𝟔𝟐
𝒔𝒊𝒏 𝑩 =
𝟕𝟐(𝐬𝐢𝐧 𝟑𝟎)
𝟔𝟐
𝑩 = 𝒔𝒊𝒏−𝟏
𝟕𝟐(𝐬𝐢𝐧 𝟑𝟎)
𝟔𝟐
𝑩 = 𝟑𝟓. 𝟓°
∠B cannot be 35.5° since
it is an obtuse angle.

LAW OF SINES
𝟑𝟎°
𝟕𝟐𝒎
𝟔𝟐𝒎
?
Solution:
𝑨 𝑩
𝑪
𝑩′
∠𝑩𝑩′𝑪 = 𝟑𝟓. 𝟓°
𝑩𝑪 ≅ 𝑩′𝑪
∠𝑩𝑩′𝑪 ≅ ∠𝑩′𝑩𝑪
∠𝑩′𝑩𝑪 = 𝟑𝟓. 𝟓°
ΔBB’C is an isosceles
triangle
𝟔𝟐𝒎 ∠𝑨𝑩𝑪 = 𝟏𝟖𝟎° − 𝟑𝟓. 𝟓°
∠𝑩′𝑩𝑪 + ∠𝑨𝑩𝑪 = 𝟏𝟖𝟎°
They are
supplementary
∠𝑨𝑩𝑪 = 𝟏𝟒𝟒. 𝟓°
𝑩 = 𝟑𝟓. 𝟓°
∠𝑨 + ∠𝑩 + ∠𝑪 = 𝟏𝟖𝟎°
𝟑𝟎° + 𝟏𝟒𝟒. 𝟓° + ∠𝑪 = 𝟏𝟖𝟎°
∠𝑪 = 𝟓. 𝟓°

LAW OF SINES
𝟑𝟎°
𝟕𝟐𝒎
𝟔𝟐𝒎
?
Solution:
𝑨 𝑩
𝑪
𝟔𝟐𝒎
∠𝑪 = 𝟓. 𝟓°
𝒄
𝒔𝒊𝒏 𝑪
=
𝒂
𝒔𝒊𝒏 𝑨
𝒄
𝒔𝒊𝒏 𝟓. 𝟓
=
𝟔𝟐
𝒔𝒊𝒏 𝟑𝟎
𝒄 =
𝟔𝟐(𝒔𝒊𝒏 𝟓. 𝟓)
𝒔𝒊𝒏 𝟑𝟎
𝒄 = 𝟏𝟐 𝒎
𝟏𝟐 𝒎
The bases of the cables are 12 m apart.
𝟓. 𝟓°

LAW OF COSINES
LAW OF COSINES
A
C
B
a
b
c
h
x 𝑐 − 𝑥
D

LAW OF COSINES
LAW OF COSINES
Consider ΔADC
A
C
B
a
b
c
h
x 𝑐 − 𝑥
D
+ =
Consider ΔBDC
𝑐2 − 2𝑐𝑥 + 𝑥2 + ℎ2 = 𝑎2
𝑐2
− 2𝑐𝑥 + 𝑏2
= 𝑎2
𝑎2 = 𝑏2 + 𝑐2 −2𝑐𝑥
cos 𝐴 =
𝑥
𝑏
; 𝑥 = 𝑏 cos 𝐴
𝑎2
= 𝑏2
+ 𝑐2
−2𝑐(𝑏𝑐𝑜𝑠 𝐴)
𝑎2 = 𝑏2 + 𝑐2 −2𝑏𝑐 𝑐𝑜𝑠 𝐴
𝑯𝒆𝒏𝒄𝒆,
𝑥2
𝑏2
ℎ2
(𝑐 − 𝑥)2
ℎ2 𝑎2
+ =

LAW OF COSINES
LAW OF COSINES
A
C
B
a
b
c
𝑎2
= 𝑏2
+ 𝑐2
−2𝑏𝑐 𝑐𝑜𝑠 𝐴
𝑏2
= 𝑎2
+ 𝑐2
−2𝑎𝑐 𝑐𝑜𝑠 𝐵
𝑐2
= 𝑎2
+ 𝑏2
−2𝑎𝑏 𝑐𝑜𝑠 𝐶

LAW OF COSINES
LAW OF COSINES
𝑎2
= 𝑏2
+ 𝑐2
𝑏2
= 𝑎2
+ 𝑐2
𝑐2
= 𝑎2
+ 𝑏2
Conditions:
1. Given two sides and an included angle (SAS);
2. Given three sides (SSS).

LAW OF COSINES
30°
A C
B
?
𝑐2 = 𝑎2 + 𝑏2 −2𝑎𝑏 𝑐𝑜𝑠 𝐶
Solution:
𝒄 ≈ 𝟓. 𝟐𝟕𝒄𝒎
b = 10cm
𝑐2 = (7)2+(10)2 −2(7)(10) 𝑐𝑜𝑠 30
𝑐2 = 49 + 100 − 140 (0.866)
𝑐2 = 27.76
𝑐 = 27.76
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Code: 951385

LAW OF COSINES
A C
B
?
Solution:
b = 15cm
𝑏2
= 𝑎2
+ 𝑐2
𝑐𝑜𝑠 𝐵 =
𝑎2
+ 𝑐2
− 𝑏2
2𝑎𝑐
𝑐𝑜𝑠 𝐵 =
(21)2
+(28)2
− (15)2
2(21)(28)
𝑐𝑜𝑠 𝐵 ≈ 0.8503
𝐵 ≈ 𝑐𝑜𝑠−1
0.8503
𝐵 ≈ 31.76°
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Code: 623431

Lets Get Real:
LAW OF COSINES
A plane is 1 km from one landmark and 2 km from
another. From the planes point of view the land
between them subtends an angle of 45°. How far
apart are the landmarks?
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Code: 9923688

LAW OF COSINES
(P)
R
45°
p=?
Solution:
Q
r =1 km
𝑝2
= 𝑞2
+ 𝑟2
−2𝑞𝑟 𝑐𝑜𝑠 𝑃
𝑝2
= 22
+ 12
−2(2)(1) 𝑐𝑜𝑠 45
𝑝2
= 4 + 1 − 4 𝑐𝑜𝑠 45
𝑝2
= 4 + 1 − 4 (0.7071)
𝑝2
= 2.17
𝑝 = 2.17
𝑝 ≈ 1.47 𝑘𝑚
∴, 𝑡ℎ𝑒 𝑙𝑎𝑛𝑑𝑚𝑎𝑟𝑘𝑠 𝑎𝑟𝑒 1.47 𝑘𝑚 𝑎𝑝𝑎𝑟𝑡.

LAW OF SINES
OBLIQUE TRIANGLES
𝐬𝐢𝐧 𝑨
𝒂
=
𝐬𝐢𝐧 𝑩
𝒃
=
𝐬𝐢𝐧 𝑪
𝒄
𝒂
𝐬𝐢𝐧 𝑨
=
𝒃
𝐬𝐢𝐧 𝑩
=
𝒄
𝐬𝐢𝐧 𝑪
𝒐𝒓
𝑎2
= 𝑏2
+ 𝑐2
𝑏2
= 𝑎2
+ 𝑐2
𝑐2
= 𝑎2
+ 𝑏2
LAW OF COSINES

Let’s do the exercises!
OBLIQUE TRAINGLES
1. Find the m∠x to nearest whole degree.
118°
12
25
x
35
Answer: m∠x = 39

OBLIQUE TRAINGLES
2. Find the m∠x to nearest whole degree.
120°
35 x
30
Answer: m∠x = 132

OBLIQUE TRAINGLES
3. Find the measures of A, B and C to nearest whole degree.
Answer: m∠A = 60
m∠B = 75
m∠C = 45

OBLIQUE TRAINGLES
4. Find the value of m∠ABD to nearest whole degree.
Answer: No solution

OBLIQUE TRAINGLES
Answer: 13.6ft
5. A monopole is supported by wires attached to a common
base. One of the wires attached to its top is 95ft long
and has an angle of depression 50 degrees. While
another 85ft wire is attached along the pole, how far
apart are the two wires to the nearest tenth feet?

OBLIQUE TRIANGLES
•Tey bong salamat..
•Maraming salamat po…
•Thank you..

4. From the illustration, how high is the building?
OBLIQUE TRAINGLES
A B
C
h
a
b
c
∠𝐶 = 180° − 25° − 60°
∠𝐶 = 95°
𝑏
sin 60
=
𝑐
sin 95
𝑏 =
1(sin 60)
sin 95
sin 25 =
ℎ
𝑏
ℎ = 𝑏 sin 25
ℎ =
sin 60
sin 95
(sin 25)
∴, ℎ =
sin 60 sin 25
sin 95
𝑏 =
sin 60
sin 95

OBLIQUE TRIANGLES .pptx

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OBLIQUE TRIANGLES .pptx