Question 1202430
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I'll do problem 1 to get you started.


The phrasing "solve the triangle" means "find the missing sides and missing angles".


Side a is opposite angle A.
Side b is opposite angle B.
Side c is opposite angle C.


The right angle, aka 90 degree angle, is at C
C = 90
We denote this with a square angle marker as shown in the diagrams below.


Angles A and B are complementary. They add to 90
A+B = 90
76+B = 90
B = 90-76
B = 14


Or we could say:
A+B+C = 180
76+B+90 = 180
166+B = 180
B = 180-166
B = 14


This is what we have so far
{{{
drawing(400,400,-5,5,-5,5,
line(-3,-3,3,-3),
line(3,-3,3,3),
line(3,3,-3,-3),
line(3-0.5,-3,3-0.5,-3+0.5),
line(3-0.5,-3+0.5,3,-3+0.5),

circle(-3,-3,0.04),
circle(-3,-3,0.06),
circle(-3,-3,0.08),
circle(-3,-3,0.10),

circle(3,-3,0.04),
circle(3,-3,0.06),
circle(3,-3,0.08),
circle(3,-3,0.10),

circle(3,3,0.04),
circle(3,3,0.06),
circle(3,3,0.08),
circle(3,3,0.10),

locate(3+0.1,0,"a=13"),
locate(-0.3,-3-0.1,"b"),
locate(-0.4,0.4,"c"),

locate(-3-0.4,-3+0.2,"A"),
locate(3+0.2,3+0.2,"B"),
locate(3+0.2,-3+0.2,"C"),

locate(-3+0.5,-3+0.65,76^o),
locate(3-0.65,3-0.5,14^o),

locate(-4.5,-4,matrix(1,2,"Diagram","not")),
locate(-4.5,-4.5,matrix(1,2,"to","scale"))
)
}}}


We'll use the soh cah toa trig ratios to fill in the missing sides.
soh = sine opposite hypotenuse
cah = cosine adjacent hypotenuse
toa = tangent opposite adjacent


Here is one way to find side c.
sin(angle) = opposite/hypotenuse
sin(A) = a/c
sin(76) = 13/c
c*sin(76) = 13
c = 13/sin(76)
c = 13.39797718
c = 13.398 which is approximate
Make sure your calculator is in degree mode.
I'll let the student explore the 2nd pathway to determine c (hint: cos(B) = a/c).


Let's find side b.
sin(angle) = opposite/hypotenuse
sin(B) = b/c
sin(14) = b/13.39797718
b = 13.39797718*sin(14)
b = 3.24126404
b = 3.241
There are three more pathways using a trig ratio, which I'll let the student explore further (hints: cos(A) = b/c, tan(A) = a/b, and tan(B) = b/a)


Or we could use the pythagorean theorem
a^2+b^2 = c^2
13^2 + b^2 = (13.39797718)^2
169 + b^2 = 179.5057925158
b^2 = 179.5057925158 - 169
b^2 = 10.5057925158
b = sqrt(10.5057925158)
b = 3.24126403055969
b = 3.241


Here's the completed right triangle with all angles and all sides labeled. The decimal values are approximate. Round however your teacher instructs.
{{{
drawing(400,400,-5,5,-5,5,
line(-3,-3,3,-3),
line(3,-3,3,3),
line(3,3,-3,-3),
line(3-0.5,-3,3-0.5,-3+0.5),
line(3-0.5,-3+0.5,3,-3+0.5),

circle(-3,-3,0.04),
circle(-3,-3,0.06),
circle(-3,-3,0.08),
circle(-3,-3,0.10),

circle(3,-3,0.04),
circle(3,-3,0.06),
circle(3,-3,0.08),
circle(3,-3,0.10),

circle(3,3,0.04),
circle(3,3,0.06),
circle(3,3,0.08),
circle(3,3,0.10),


locate(-3-0.4,-3+0.2,"A"),
locate(3+0.2,3+0.2,"B"),
locate(3+0.2,-3+0.2,"C"),

locate(3+0.1,0,"a=13"),
locate(-0.5,-3-0.2,"b=3.241"),
locate(-1.4,0.6,"c=13.398"),

locate(-3+0.5,-3+0.65,76^o),
locate(3-0.65,3-0.5,14^o),

locate(-4.5,-4,matrix(1,2,"Diagram","not")),
locate(-4.5,-4.5,matrix(1,2,"to","scale"))
)
}}}
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