Download Solving Triangles: Assignment VIII in Mathematics 1B and more Assignments Pre-Calculus in PDF only on Docsity!
Mathematics 1B
Assignment VIII
α β
γ (^) a b
c
d
h
α β
(a) (b)
Figure 1
- Solve triangle(s) using the information given and the triangle labeling in the Figure 1-(a)
(1) α = 52o, γ = 47o, a = 13 centimeters (2) α = 38. 9 o, a = 30.0 inches, b = 42.7 inches (Hint: has two triangles) (3) β = 29o 30 ′, a = 43.2 millimeters, b = 56.5 millimeters
Solution. (1). Solve for β:
α + β + γ = 180 o β = 180 o^ − (α + γ) = 1800 − (52o^ + 47o) = 81 o
Solve for b:
sin α a
= sin^ β b b =
a sin β sin α = 13 sin 81
o sin 52o = 16 cm 1
Solve for c:
sin α a
= sin^ γ c c =
a sin γ sin α = 13 sin 47
o sin 52o = 12 cm
(2). There are two possible triangles I and II (Figure 2).
A B
D
A C
D
c (^) c′
42 .7 in (^30) .0 in^42 .7 in^30 .0 in
- 9 o^ β
γ
- 9 o^ β
′
γ′
I II
Figure 2
Solve for β and β′^ We start by finding β and β′^ using the law of sines:
sin β b
= sin^ α a sin β = b^ sin^ α a
=^42 .7 sin 38.^9
o
- 0
Angle β can be either obtuse or acute:
β = 180 o^ − sin−^1 0. 893801 or β′^ = sin−^1 0. 893801 = 180 o^ − 63. 4 o^ = 116. 6 o^ = 63. 4 o
Solve for γ and γ′^ We next find γ and γ′:
γ = 180 o^ − (38. 9 o^ + 116. 6 o) = 24. 5 o γ′^ = 180 o^ − (38. 9 o^ + 63. 4 o) = 77. 7 o
- In the Figure 1-(b), show that (This formula can be used to measure the height of the object, for example, refer Ex. 7-1: 39)
h = d sin sin(^ αβ^ sin− α^ β)
Solution. Redrawing the figure and labeling (Figure 4), we have: Triangle
d
h
α β
A
F E C
γ
a
Figure 4
ACE is a right triangle, hence h a
= sin β
and h = a sin β. Triangle AEF is not a right triangle, from the law of sines, we have sin γ d
= sin^ α a hence
a =
d sin α sin γ. Thus, we have
h = a sin β = d^ sin^ α sin γ
sin β = d sin^ α^ sin^ β sin γ Finally, since β is an exterior angle of the triangle AEF, and the exterior angle of a triangle has measure equal to the sum of the two nonadjacent interior angles. Hence
α + γ = β γ = β − α
Hence, we have
h = d sin^ α^ sin^ β sin(β − α) §
