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6 September, 03:33

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

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  1. 6 September, 04:14
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    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°.
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