Question

In ΔАВC, AB = 4 cm, BC = 5 cm and CA = 6 cm. Find the longest angle of ΔABC.

Answer 1

Given:

• Triangle ABC with AB = 4 cm, BC = 5 cm, CA = 6 cm.

Step-wise calculation:

1. The largest side is CA = 6.

So, the largest angle is the angle opposite it, i.e., ∠B.

2. Use the Law of Cosines to find ∠B side opposite is AC = 6; adjacent sides AB = 4 and BC = 5:

`cos B = (AB^2 + BC^2 - AC^2)/(2 xx AB xx BC)` 

= `(4^2 + 5^2 - 6^2)/(2 xx 4 xx 5)` 

= `(16 + 25 - 36)/40` 

= `5/40` 

= `1/8`

= 0.125

Therefore, B = arc cos (0.125) = 82.82°.

3. For completeness,

`cos A = (AB^2 + AC^2 - BC^2)/(2 xx AB xx AC)` 

= `27/48`

= 0.5625 

⇒ A = 55.77°

`cos C = (BC^2 + AC^2 - AB^2)/(2 xx BC xx AC)` 

= `45/60`

= 0.75

⇒ C = 41.41°

Sum

= 82.82° + 55.77° + 41.41°

= 180°

The longest angle of ΔABC is ∠B = 82.82°.