# In triangle ABC the height AD is drawn, from point D the perpendicular DN is lowered to the side AC.

In triangle ABC the height AD is drawn, from point D the perpendicular DN is lowered to the side AC. DN – 4 cm, BD – 2, angle C – a. Find the area of triangle ABC.

In a right-angled triangle СDН, we determine the length of the hypotenuse СD.

Sinα = DH / CD.

СD = DH / Sinα = 4 / Sinα see.

Angle СDН = (90 – α), then in triangle АDН the angle АDН = (90 – (90 – α) = α.

Then Cosα = DH / AD.

AD = DH / Cosα = 4 / Cosα.

Segment ВС = ВD + СD = 2 + 4 / Sinα.

The area of triangle ABC is equal to: Sас = ВС * АD / 2 = (2 + 4 / Sinα) * (4 / Cosα) / 2 = (8/2 * Cosα) + (16/2 * Cosα * Sinα) = (8 + 4 * Sinα) / (Sinα * Cosα).

Answer: The area of triangle ABC is equal to (8 + 4 * Sinα) / (Sinα * Cosα) cm2.

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