3. At the end of the session, teachers should be able to:
โข Discuss the nature of oblique triangle;
โข Derive the Law of Sines and Cosines
โข Solve oblique triangle related problems applying the law of
sines or cosines
OBLIQUE TRIANGLES
DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
OBJECTIVES:
4. 1.Which triangle DOES NOT belong to the group?
OBLIQUE TRAINGLES
DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
a
b
d
c
5. 2. Which of the following is NOT a right triangle?
OBLIQUE TRAINGLES
DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
a
b
d
c
6. 3. Triangle ABC is defined by the following measures:
โ A = 30ยฐ, b = 10. What value of side a would lead to
more than one possible triangles?
OBLIQUE TRAINGLES
DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
a. a < 10 c. 10sin30 < a < 10
b. a โฅ 10 sin 30 d. 10 sin 30 โฅ a โฅ 10
7. 4. From the illustration, how high is the building?
OBLIQUE TRAINGLES
DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
a.
b.
d.
c.
โ =
sin 95 sin 60
sin 25
๐๐๐ ๐ข๐๐๐๐๐๐๐๐ก ๐๐๐๐๐๐๐๐ก๐๐๐
โ =
sin 25 sin 60
sin 95
โ =
sin 95
sin 25 sin 60
8. OBLIQUE TRIANGLE
OBLIQUE TRAINGLES
DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
- is a triangle containing NO 90ยฐ angle.
Examples:
- Any triangle which is NOT a right triangle.
9. LAW OF SINES
LAW OF SINES
DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
A
C
B
a
b
c
h
10. B
a
LAW OF SINES
LAW OF SINES
DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
A
C
b
c
h
๐ฌ๐ข๐ง ๐จ =
๐
๐
๐ ๐ฌ๐ข๐ง ๐จ =
๐
๐
๐ ๐ฌ๐ข๐ง ๐จ = ๐
11. B
a
LAW OF SINES
LAW OF SINES
DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
A
C
b
c
h
๐ฌ๐ข๐ง ๐ฉ =
๐
๐
๐ ๐ฌ๐ข๐ง ๐ฉ =
๐
๐
๐ ๐ฌ๐ข๐ง ๐ฉ = ๐
12. LAW OF SINES
LAW OF SINES
DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
๐ฌ๐ข๐ง ๐ฉ =
๐
๐
๐ ๐ฌ๐ข๐ง ๐ฉ =
๐
๐
๐ ๐ฌ๐ข๐ง ๐ฉ = ๐
B
a
C
h
c
A
C
B
a
b
c
h
13. LAW OF SINES
LAW OF SINES
DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
A
C
B
a
b
c
h
๐ฌ๐ข๐ง ๐จ =
๐
๐
๐ ๐ฌ๐ข๐ง ๐จ =
๐
๐
๐ ๐ฌ๐ข๐ง ๐จ = ๐
๐ฌ๐ข๐ง ๐ฉ =
๐
๐
๐ ๐ฌ๐ข๐ง ๐ฉ =
๐
๐
๐ ๐ฌ๐ข๐ง ๐ฉ = ๐
14. LAW OF SINES
LAW OF SINES
DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
A
C
B
a
b
c
h
๐ ๐ฌ๐ข๐ง ๐จ = ๐ ๐ ๐ฌ๐ข๐ง ๐ฉ = ๐
๐ ๐ฌ๐ข๐ง ๐จ = ๐ = ๐ ๐ฌ๐ข๐ง ๐ฉ
๐ ๐ฌ๐ข๐ง ๐จ = ๐ ๐ฌ๐ข๐ง ๐ฉ
๐
๐ฌ๐ข๐ง ๐จ
=
๐
๐ฌ๐ข๐ง ๐ฉ
=
๐
๐ฌ๐ข๐ง ๐ช
๐
๐ฌ๐ข๐ง ๐จ
=
๐
๐ฌ๐ข๐ง ๐ฉ
๐ฏ๐๐๐๐,
15. LAW OF SINES
LAW OF SINES
DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
๐ฌ๐ข๐ง ๐จ
๐
=
๐ฌ๐ข๐ง ๐ฉ
๐
=
๐ฌ๐ข๐ง ๐ช
๐
Conditions:
1. Given two angles and a side (AAS).
2. Given two sides and an angle (SSA);
๐
๐ฌ๐ข๐ง ๐จ
=
๐
๐ฌ๐ข๐ง ๐ฉ
=
๐
๐ฌ๐ข๐ง ๐ช
๐๐
16. Illustrative Example 1:
LAW OF SINES
DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
25ยฐ
P
R
Q p = ?
๐
๐๐๐ ๐ท
=
๐
๐๐๐ ๐ธ
Solution:
๐
๐๐๐ ๐๐
=
๐๐
๐๐๐ ๐๐๐
(๐๐๐ ๐๐)
๐
๐๐๐ ๐๐
=
๐๐
๐๐๐ ๐๐๐
๐ =
๐๐ (๐๐๐ ๐๐)
๐๐๐ ๐๐๐
๐ โ ๐. ๐๐๐๐
115ยฐ
17. Illustrative Example 2:
LAW OF SINES
DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
36ยฐ
A C
B
?
๐ฌ๐ข๐ง ๐ช
๐
=
๐ฌ๐ข๐ง ๐จ
๐
Solution:
๐ฌ๐ข๐ง ๐ช
๐๐
=
๐ฌ๐ข๐ง ๐๐
๐๐
๐๐
๐ฌ๐ข๐ง ๐ช
๐๐
=
๐ฌ๐ข๐ง ๐๐
๐๐
๐ฌ๐ข๐ง ๐ช =
๐๐(๐ฌ๐ข๐ง ๐๐)
๐๐
๐ฌ๐ข๐ง ๐ช = ๐. ๐๐๐๐
๐ช = ๐๐๐โ๐(๐. ๐๐๐๐)
๐ช โ ๐๐. ๐๐ยฐ
18. Lets Get Real:
LAW OF SINES
DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
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?
19. Illustrating the problem:
LAW OF SINES
DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
2.5 km
Juan
(J)
Romella
(R)
(S)
55ยฐ 65ยฐ
r=?
Solution:
โ ๐บ = ๐๐๐ยฐ โ ๐๐ยฐ โ ๐๐ยฐ
โ ๐บ = ๐๐ยฐ
๐
๐๐๐ ๐น
=
๐
๐๐๐ ๐บ
๐
๐๐๐ ๐๐
=
๐. ๐
๐๐๐ ๐๐
(๐๐๐ ๐๐)
๐
๐๐๐ ๐๐
=
๐. ๐
๐๐๐ ๐๐
๐ =
๐. ๐ (๐๐๐ ๐๐)
๐๐๐ ๐๐
๐ โ ๐. ๐ ๐๐
โด, ๐๐๐ ๐๐๐๐ ๐๐ ๐. ๐ ๐๐ ๐๐๐๐ ๐๐๐๐ ๐ฑ๐๐๐.
20. Lets Try:
LAW OF SINES
DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
Find the measure of angle B in a triangle ABC given that
โ A = 65ยฐ a = 15 cm and b = 20 cm.
๐๐๐ ๐ฉ
๐
=
๐๐๐ ๐จ
๐
๐๐๐ ๐ฉ
๐๐
=
๐๐๐ ๐๐
๐๐
๐๐๐ ๐ฉ =
๐๐(๐๐๐ ๐๐)
๐๐
๐๐๐ ๐ฉ =
๐๐(๐๐๐ ๐๐)
๐๐
๐๐๐ ๐ฉ = ๐. ๐๐๐๐
๐ฉ = ๐๐๐โ๐
(๐. ๐๐๐๐)
๐ฌ๐น๐น๐ถ๐น
๐ป๐๐ ๐๐๐๐๐๐๐๐ ๐ซ๐ถ๐ฌ๐บ ๐ต๐ถ๐ป ๐๐๐๐๐
๐๐๐๐๐ ๐๐ ๐๐๐ ๐๐๐๐๐ ๐๐๐๐ ๐๐๐๐๐
21. Law of Sines โ Ambiguous Case:
LAW OF SINES
DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
The ambiguous case can occur when we are given the angle โ side โ
side. Consider a triangle in which you are given a, b, and A. (๐ = ๐ ๐๐๐ ๐จ)
A is acuteโฆ
A
b
a
h
Necessary condition
a < h
Triangles possible
None
22. Law of Sines โ Ambiguous Case:
LAW OF SINES
DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
The ambiguous case can occur when we are given the angle โ side โ
side. Consider a triangle in which you are given a, b, and A. (๐ = ๐ ๐๐๐ ๐จ)
A is acuteโฆ
A
b a
h
Necessary condition
a = h
Triangles possible
One
23. Law of Sines โ Ambiguous Case:
LAW OF SINES
DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
The ambiguous case can occur when we are given the angle โ side โ
side. Consider a triangle in which you are given a, b, and A. (๐ = ๐ ๐๐๐ ๐จ)
A is acuteโฆ
A
b a
h
Necessary condition
a โฅ b
Triangles possible
One
24. Law of Sines โ Ambiguous Case:
LAW OF SINES
DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
The ambiguous case can occur when we are given the angle โ side โ
side. Consider a triangle in which you are given a, b, and A. (๐ = ๐ ๐๐๐ ๐จ)
A is acuteโฆ
A
b a
h
a
Necessary condition
h < a < b
Triangles possible
Two
25. Law of Sines โ Ambiguous Case:
LAW OF SINES
DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
The ambiguous case can occur when we are given the angle โ side โ
side. Consider a triangle in which you are given a, b, and A. (๐ = ๐ ๐๐๐ ๐จ)
A is obtuseโฆ
A
b
a
Necessary condition
a โค b
Triangles possible
None
26. Law of Sines โ Ambiguous Case:
LAW OF SINES
DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
The ambiguous case can occur when we are given the angle โ side โ
side. Consider a triangle in which you are given a, b, and A. (๐ = ๐ ๐๐๐ ๐จ)
A is obtuseโฆ
A
b
a
Necessary condition
a > b
Triangles possible
One
27. DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
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)?
28. DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
LAW OF SINES
Illustrating the problem:
๐๐ยฐ
๐๐๐
๐๐๐
?
Solution:
๐จ ๐ฉ
๐ช
We need to find mโ ABC in
order to obtain the mโ ACB.
๐ช๐๐๐๐ ๐๐๐ ๐๐๐๐๐๐๐๐๐
๐ = ๐๐(๐๐๐ ๐๐)
๐ = ๐๐๐
๐บ๐๐๐๐ ๐ < ๐ < ๐, ๐๐๐๐๐ ๐๐๐
๐๐๐ ๐๐๐๐๐๐๐๐ ๐๐๐๐๐๐๐๐๐
Angle โ side โ side
๐
29. DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
LAW OF SINES
Illustrating the problem:
๐๐ยฐ
๐๐๐
๐๐๐
?
Solution:
๐จ ๐ฉ
๐ช
๐ฌ๐ข๐ง ๐ฉ
๐
=
๐ฌ๐ข๐ง ๐จ
๐
๐ฌ๐ข๐ง ๐ฉ
๐๐
=
๐ฌ๐ข๐ง ๐๐
๐๐
๐๐๐ ๐ฉ =
๐๐(๐ฌ๐ข๐ง ๐๐)
๐๐
๐ฉ = ๐๐๐โ๐
๐๐(๐ฌ๐ข๐ง ๐๐)
๐๐
๐ฉ = ๐๐. ๐ยฐ
โ B cannot be 35.5ยฐ since
it is an obtuse angle.
30. DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
LAW OF SINES
Illustrating the problem:
๐๐ยฐ
๐๐๐
๐๐๐
?
Solution:
๐จ ๐ฉ
๐ช
๐ฉโฒ
โ ๐ฉ๐ฉโฒ๐ช = ๐๐. ๐ยฐ
๐ฉ๐ช โ ๐ฉโฒ๐ช
โ ๐ฉ๐ฉโฒ๐ช โ โ ๐ฉโฒ๐ฉ๐ช
โ ๐ฉโฒ๐ฉ๐ช = ๐๐. ๐ยฐ
ฮBBโC is an isosceles
triangle
๐๐๐ โ ๐จ๐ฉ๐ช = ๐๐๐ยฐ โ ๐๐. ๐ยฐ
โ ๐ฉโฒ๐ฉ๐ช + โ ๐จ๐ฉ๐ช = ๐๐๐ยฐ
They are
supplementary
โ ๐จ๐ฉ๐ช = ๐๐๐. ๐ยฐ
๐ฉ = ๐๐. ๐ยฐ
โ ๐จ + โ ๐ฉ + โ ๐ช = ๐๐๐ยฐ
๐๐ยฐ + ๐๐๐. ๐ยฐ + โ ๐ช = ๐๐๐ยฐ
โ ๐ช = ๐. ๐ยฐ
31. DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
LAW OF SINES
Illustrating the problem:
๐๐ยฐ
๐๐๐
๐๐๐
?
Solution:
๐จ ๐ฉ
๐ช
๐๐๐
โ ๐ช = ๐. ๐ยฐ
๐
๐๐๐ ๐ช
=
๐
๐๐๐ ๐จ
๐
๐๐๐ ๐. ๐
=
๐๐
๐๐๐ ๐๐
๐ =
๐๐(๐๐๐ ๐. ๐)
๐๐๐ ๐๐
๐ = ๐๐ ๐
๐๐ ๐
The bases of the cables are 12 m apart.
๐. ๐ยฐ
32. LAW OF COSINES
LAW OF COSINES
DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
A
C
B
a
b
c
h
x ๐ โ ๐ฅ
D
33. LAW OF COSINES
LAW OF COSINES
DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
A
C
B
a
b
c
h
x ๐ โ ๐ฅ
D
34. LAW OF COSINES
LAW OF COSINES
DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
Consider ฮADC
A
C
B
a
b
c
h
x ๐ โ ๐ฅ
D
+ =
Consider ฮBDC
๐2 โ 2๐๐ฅ + ๐ฅ2 + โ2 = ๐2
๐2
โ 2๐๐ฅ + ๐2
= ๐2
๐2 = ๐2 + ๐2 โ2๐๐ฅ
cos ๐ด =
๐ฅ
๐
; ๐ฅ = ๐ cos ๐ด
๐2
= ๐2
+ ๐2
โ2๐(๐๐๐๐ ๐ด)
๐2 = ๐2 + ๐2 โ2๐๐ ๐๐๐ ๐ด
๐ฏ๐๐๐๐,
๐ฅ2
๐2
โ2
(๐ โ ๐ฅ)2
โ2 ๐2
+ =
35. LAW OF COSINES
LAW OF COSINES
DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
A
C
B
a
b
c
๐2
= ๐2
+ ๐2
โ2๐๐ ๐๐๐ ๐ด
๐2
= ๐2
+ ๐2
โ2๐๐ ๐๐๐ ๐ต
๐2
= ๐2
+ ๐2
โ2๐๐ ๐๐๐ ๐ถ
36. LAW OF COSINES
LAW OF COSINES
DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
๐2
= ๐2
+ ๐2
โ2๐๐ ๐๐๐ ๐ด
๐2
= ๐2
+ ๐2
โ2๐๐ ๐๐๐ ๐ต
๐2
= ๐2
+ ๐2
โ2๐๐ ๐๐๐ ๐ถ
Conditions:
1. Given two sides and an included angle (SAS);
2. Given three sides (SSS).
37. Illustrative Example 1:
LAW OF COSINES
DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
30ยฐ
A C
B
?
๐2 = ๐2 + ๐2 โ2๐๐ ๐๐๐ ๐ถ
Solution:
๐ โ ๐. ๐๐๐๐
b = 10cm
๐2 = (7)2+(10)2 โ2(7)(10) ๐๐๐ 30
๐2 = 49 + 100 โ 140 (0.866)
๐2 = 27.76
๐ = 27.76
38. Illustrative Example 2:
LAW OF COSINES
DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
A C
B
?
Solution:
b = 15cm
๐2
= ๐2
+ ๐2
โ2๐๐ ๐๐๐ ๐ต
๐๐๐ ๐ต =
๐2
+ ๐2
โ ๐2
2๐๐
๐๐๐ ๐ต =
(21)2
+(28)2
โ (15)2
2(21)(28)
๐๐๐ ๐ต โ 0.8503
๐ต โ ๐๐๐ โ1
0.8503
๐ต โ 31.76ยฐ
39. Lets Get Real:
LAW OF COSINES
DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
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 45ยฐ. How far
apart are the landmarks?
40. Illustrating the problem:
LAW OF COSINES
DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
(P)
R
45ยฐ
p=?
Solution:
Q
r =1 km
๐2
= ๐2
+ ๐2
โ2๐๐ ๐๐๐ ๐
๐2
= 22
+ 12
โ2(2)(1) ๐๐๐ 45
๐2
= 4 + 1 โ 4 ๐๐๐ 45
๐2
= 4 + 1 โ 4 (0.7071)
๐2
= 2.17
๐ = 2.17
๐ โ 1.47 ๐๐
โด, ๐กโ๐ ๐๐๐๐๐๐๐๐๐ ๐๐๐ 1.47 ๐๐ ๐๐๐๐๐ก.
41. LAW OF SINES
OBLIQUE TRIANGLES
DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
๐ฌ๐ข๐ง ๐จ
๐
=
๐ฌ๐ข๐ง ๐ฉ
๐
=
๐ฌ๐ข๐ง ๐ช
๐
๐
๐ฌ๐ข๐ง ๐จ
=
๐
๐ฌ๐ข๐ง ๐ฉ
=
๐
๐ฌ๐ข๐ง ๐ช
๐๐
๐2
= ๐2
+ ๐2
โ2๐๐ ๐๐๐ ๐ด
๐2
= ๐2
+ ๐2
โ2๐๐ ๐๐๐ ๐ต
๐2
= ๐2
+ ๐2
โ2๐๐ ๐๐๐ ๐ถ
LAW OF COSINES
42. Letโs do the exercises!
OBLIQUE TRAINGLES
DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
1. Find the mโ x to nearest whole degree.
118ยฐ
12
25
x
35
Answer: mโ x = 39
Y
Y =180 - 118
Y = 62ห
SinX
x
๏ฝ
SinY
y
SinX
25
Sin62
35
๏ฝ
35
62
25Sin
SinX ๏ฝ
63068
.
๏ฝ
SinX
39
๏ฝ
X
43. Letโs do the exercises!
OBLIQUE TRAINGLES
DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
2. Find the mโ x to nearest whole degree.
120ยฐ
35 x
30
Answer: mโ x = 132
SinY
Sin
30
120
35
๏ฝ
Y
35
120
30Sin
SinY ๏ฝ
๏ฏ
48
๏ฝ
Y
180
๏ฝ
๏ซ X
Y
48
180 ๏ญ
๏ฝ
X
๏ฏ
132
๏ฝ
X
44. Letโs do the exercises!
OBLIQUE TRAINGLES
DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
3. Find the measures of A, B and C to nearest whole degree.
Answer: mโ A = 60
mโ B = 75
mโ C = 45
a2 = b2 + c2 โ 2bcCosA
172 = 192 + 142 โ 2(19)(14)CosA
289= 361 + 196 โ 532CosA
-268=-532CosA
CosA = .50376
A = 60ห
b2 = a2 + c2 โ 2acCosB
192 = 172 + 142 โ 2(17)(14)CosB
361= 289 + 196 โ 476CosB
-125=-476CosB
CosA = .26050
A = 75ห
c2 = a2 + b2 โ 2abCosC
142 = 172 + 192 โ 2(17)(19)CosC
196= 289 + 361 โ 646CosC
-454=-646CosC
CosA = .70279
A = 45ห
45. Letโs do the exercises!
OBLIQUE TRAINGLES
DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
4. Find the value of mโ ABD to nearest whole degree.
Answer: No solution
46. Letโs do the exercises!
OBLIQUE TRAINGLES
DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
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?
47. DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
OBLIQUE TRIANGLES
โขTey bong salamat..
โขMaraming salamat poโฆ
โขThank you..
48. 4. From the illustration, how high is the building?
OBLIQUE TRAINGLES
DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
A B
C
h
a
b
c
โ ๐ถ = 180ยฐ โ 25ยฐ โ 60ยฐ
โ ๐ถ = 95ยฐ
๐
sin 60
=
๐
sin 95
๐ =
1(sin 60)
sin 95
sin 25 =
โ
๐
โ = ๐ sin 25
โ =
sin 60
sin 95
(sin 25)
โด, โ =
sin 60 sin 25
sin 95
๐ =
sin 60
sin 95