A triangle has sides with the following lengths: AB=3, BC=4, CA=5. Which angle is the smallest?

A = 36.9°

Step-by-step explanation:

In this triangle we know the three sides:

AB = 3,

BC = 4 and

CA = 5.

Use The Law of Cosines first to find angle A first:

cos A = (BC² + CA² - AB²) / 2BCCA

cos A = (4² + 5² - 3²) / (2*4*5)

cos A = (16 + 25 - 9) / 40

cos A = 0.80

A = cos⁻¹ (0.80)

A = 36.86989765°

A = 36.9° to one decimal place.

Next we will find another side. We use The Law of Cosines again, this time for angle B:

cos B = (CA² + AB² - BC²) / 2CAAB

cos B = (5² + 3² - 4²) / (2*5*3)

cos B = (25 + 9 - 16) / 30

cos B = 0.60

B = cos⁻¹ (0.60)

B = 53.13010235°

B = 53.1° to one decimal place

Finally, we can find angle C by using 'angles of a triangle add to 180°:

C = 180° - 36.86989765° - 53.13010235°

C = 90°

Now we have completely solved the triangle i. e. we have found all its angles.

So we can analyze from above that the smallest angle in the triangle ABC is A with 36.9°.