Yoder 2.

asked • 04/25/23

precalculus help

imgur.com/gallery/EhOVveC link to the question of the assignment since i cannot upload a picture here

Doug C.

Here is a Desmos graph that shows that triangle for part 1 is a right triangle. There are a few different ways to reach this conclusion. desmos.com/calculator/f5jcdvxyjx
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04/25/23

Doug C.

For part 2, find the missing 3rd angle, use Law of Sines to find another side, then area of triangle is 1/2 ab times sine of included angle)
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04/25/23

Doug C.

For part 3, this is Law of Sines ambiguous case. Sketch the triangle and try to find the altitude from vertex B to side AC (b). The altitude will equal side a, so there is only one triangle (a right triangle again).
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04/25/23

Yoder 2.

Can you give the answers please
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04/25/23

1 Expert Answer

By:

Joseph S. answered • 04/25/23

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Joe S. Experienced Math Tutor
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1.

sinC/c = sinA/a

sinC = c/a(sinA)

sinC = (4/2)(sin30°)

sinC = 1

C = 90°

(This is a 30-60-90 triangle)

B = 60°

b = a√3

b = 2√3


2.

B = 180 – (23 + 10/60) + (95 + 10/60)

B = 61 + 40/60 = 61°40’


a/sinA = b/sinB

a = b(sinA/sinB)

a = 19.1(sin(23+10/60)/sin(61 + 40/60))

a = 8.54 ft.


A = ab(sinC)/2

A = 0.5(8.54)(19.1)(sin(95 + 10/60))

A = 81.2 ft2


3.

sinB/b = sinA/a

sinB = b/a(sinA)

sinB = (15.3/7.65)(sin30°)

sinB = (15.3/7.65)(sin30°)

sinB = 1

B = 90°

(This is a 30-60-90 triangle)


C = 60°


c = a√3

c = 7.65√3

c = 13.25

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