How do you solve the right triangle given A = 28° 30' , b = 18.3?

1 Answer
Oct 25, 2015

#∠A= 28.5º, ∠B=61.5º, ∠C=90º#
#a≈4.2, b=18.3, c≈18.8#

Explanation:

We are given this triangle:

Since there are 60 minutes in a degree, we know that #∠A= 28.5º#

Since we know #∆ABC# is a right triangle, #∠C=90º#

Since there are 180º in all triangles, #∠A+∠B+∠C=180º#, which means that #28.5º+∠B+90º=180º#, which means that #∠B=61.5º#

Now our triangle is: enter image source here

Using the law of sines (#sin(A)/a=sin(B)/b#), we can obtain the length of side a: #sin(28.5)/a=sin(61.5)/18.3#

This simplifies to #sin(61.5)×a=18.3sin(28.5)#, which simplifies to #a=(18.3sin(28.5))/sin(61.5)#, which is approximately 4.2.

Now our triangle looks like this: enter image source here

From here, we can solve for side c by using the pythagorean theorem (#a^2+b^2=c^2#): #(18.3)^2+(4.2)^2=c^2#

#c=sqrt((18.3)^2+(4.2)^2)=sqrt(352.53)≈18.8#

So we end with our solved triangle that looks like this: enter image source here