If in a triangle ABC, the altitudes from the vertices A, B, C on opposite sides are in HP, then sin A, sin B, sin C are in ?

Concept:

Area of \(\rm \triangle ABC = A = \frac 12 \) × Base × Height

Let a, b and c be the sides of the triangle.

If \(\rm \dfrac {1}{a}\), \(\rm \dfrac {1}{b}\)and \(\rm \dfrac {1}{c}\) are in HP then a, b and c are in AP.

If a, b and c are in AP then 2b = a + c.

The Law of Sines says that in any given triangle, the ratio of any side length to the sine of its opposite angle is the same for all three sides of the triangle.

Calculations:

Given, in a triangle ABC, the altitudes from the vertices A, B, C on opposite sides are in HP

In triangle ABC, the altitudes from the vertices A to BC, B to AC and C to AB.

⇒ AD, BE, and CF are the altitudes and they are in HP.

⇒ Area of \(\rm \triangle ABC = A = \dfrac 12 × AD × BC\)

⇒ AD = \(\rm \dfrac {2A}{a}\)

Similarly, BE = \(\rm \dfrac {2A}{b}\) and CF = \(\rm \dfrac {2A}{c}\)

⇒ \(\rm \dfrac {2A}{a}\), \(\rm \dfrac {2A}{b}\)and \(\rm \dfrac {2A}{c}\) are in HP.

⇒ \(\rm \dfrac {1}{a}\), \(\rm \dfrac {1}{b}\)and \(\rm \dfrac {1}{c}\) are in HP.

⇒ a, b and c are in AP

\(\rm \frac {a}{sin \;A} = \frac {b}{sin \;B} = \frac {c}{sin \;C} =k\)           (sine rule)

⇒ a = k sin A, b = k sin B and c = k sin C

As we know, If a, b and c are in AP then 2b = a + c

⇒ 2k sin B = k sin A + k sin C

⇒ 2sin B = sin A + sin C

Hence sin A, sin B, sin C are in AP