How do you solve the triangle given A=83^circ20', C=54.6^circ, c=18.1?

1 Answer
Mar 16, 2018

color(white){color(black)( (m/_A=color(black)(83 1/3)^@, qquadqquad a~~22.05), (m/_B=color(black)(42 1/15)^@, qquadqquad b~~14.88), (m/_C=color(black)(54 3/5)^@, qquadqquad c = 18.1):}

Explanation:

First, convert the degrees with minutes to just degrees:

A=83^@20'

color(white)A=83^@+20'

color(white)A=83^@+20'color(blue)(*1^@/(60'))

color(white)A=83^@+20color(red)cancelcolor(black)'color(blue)(*1^@/(60color(red)cancelcolor(blue)'))

color(white)A=83^@+20^@/60

color(white)A=83^@+1^@/3

color(white)A=color(black)(83 1/3)^@

Now, we can construct a triangle based on what we know so far:

https://www.geogebra.org/geometryhttps://www.geogebra.org/geometry

We can see that we can use the law of sines to compare between angleC, c, angleA, and a (you should probably use a calculator for the last step):

color(white)=>sinA/a=sinC/c

=>sin(color(black)(83 1/3)^@)/a=sin(54.6^@)/18.1

color(white)=>a/sin(color(black)(83 1/3)^@)=18.1/sin(54.6^@)

color(white)=>a=sin(color(black)(83 1/3)^@)*18.1/sin(54.6^@)

color(white)(=>a)~~22.05

Now, since we know that the angles in a triangle must add up to 180^@, we can do some math to compute the measure of angleB:

mangleA+mangleB+mangleC=180^@

color(black)(83 1/3)^@+mangleB+54.6^@=180^@

mangleB=180^@-color(black)(83 1/3)^@-54.6^@

qquadqquad qquadqquad qquad vdots quad After some annoying work...

color(white)(mangleB)=color(black)(42 1/15)^@

We can use the law of sines once again to solve for b:

color(white)=>sinB/b=sinC/c

=>sin(color(black)(42 1/15)^@)/b=sin(54.6^@)/18.1

color(white)=>b/sin(color(black)(42 1/15)^@)=18.1/sin(54.6^@)

color(white)=>b=sin(color(black)(42 1/15)^@)*18.1/sin(54.6^@)

color(white)(=>b)~~14.88

We have solved all the parts of the triangle. Hope this helped!