In an acute-angled triangle ABC, the angle is A = 45 degrees, BC = 13cm, point D is taken on the AC side so that DC

| February 25, 2021 | education

In an acute-angled triangle ABC, the angle is A = 45 degrees, BC = 13cm, point D is taken on the AC side so that DC = 5cm, BD = -12cm. Prove that triangle BDC is right-angled, and find the area of triangle ABC.

All three measurements are given in the ВСD triangle, we apply the cosine theorem and determine the value of the ВDC angle.

ВС ^ 2 = ВD ^ 2 + СД ^ 2 – 2 * ВD * СD * CosВDC.

169 = 144 + 25 – 2 * 12 * 5 * CosВDC.

120 * CosВDC = 169 – 169 = 0.

CosВDC = 0, therefore the angle ВDC = 90, and the triangle ВDC is rectangular, which was required to be proved.

The ABD triangle is rectangular and isosceles, since the angle BAD = 45, then AD = BD = 12 cm, and the length of the segment AC = AD + CD = 12 + 5 = 17 cm.

Then Savs = AC * BD / 2 = 17 * 12/2 = 102 cm2.

Answer: The area of triangle ABC is 102 cm2.



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