Question

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

Answer 1

Answer:

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°.