Question

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

Answer 1

The smallest angle is `25.34^o`

Explanation:

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`

Answer 2

`cos A = 0.9.38`

`A = 25.33°`

Explanation:

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°`