How do you find the smallest angle in a right angled triangle whose side lengths are 6cm, 13cm and 14 cm?

2 Answers
Jun 10, 2016

The smallest angle is #25.34^o#

Explanation:

enter image source here

Since the smallest angle is always opposite the shortest side, we will solve for the value of angle A using the Law of Cosine

#a^2=b^2+c^2-2 b c CosA#

#a=6#
#b=13#
#c=14#

#6^2=13^2+14^2-2(13)(14)CosA#

#36=169+196-2(13)(14)CosA#

#36=365-364CosA#

#36-365=cancel365cancel(-365)-364CosA#

#-329=-364CosA#

#(-329)/(-364) = (cancel(-364)CosA)/cancel(-364)#

#0.9038 = CosA#

#Cos^-1(0.9038) = A#

#A=25.34^o#

Jul 30, 2016

#cos A = 0.9.38#

#A = 25.33°#

Explanation:

enter image source here

Sketch taken from Brian M's answer.

The Cosine Rule can be written in two ways: one for finding a side, and the transposed formula for finding an angle.

It Really is worth learning both Forms!

For side BC: #a^2 = b^2 + c^2 - 2 b c Cos A#

For angle A: #Cos A = (b^2 + c^2 - a^2)/(2bc)#

The smallest angle is always opposite the shortest side, so we need to find the size of Angle A in this case.

Substitute the given values:

#Cos A =(13^2 + 14^2 - 6^2)/(2xx13xx14)#

There is little point in doing long-hand calculations, or even calculating the individual answers. It it the final value that we need.
Use a calculator, making sure to divide the whole of the numerator by the whole of the denominator. (Use brackets to ensure this.)

#Cos A =((13^2 + 14^2 - 6^2))/((2xx13xx14))#

#cos A = 0.9.38#

#A = 25.33°#