In triangle ABC, AB = 15 cm, AC = 15 cm and BC = 18 cm. Find

In triangle ABC, AB = 15 cm, AC = 15 cm and BC = 18 cm. Find

(i) cos ∠ABC

(ii) sin ∠ACB.

Answer :

Here,

ABC is a triangle in which

AB = 15 cm, AC = 15 cm and BC = 18 cm

Draw AD perpendicular to BC , D is mid-point of BC.

Then, BD – DC = 9 cm

in right angled triangle ABD,

AB² = AD² + BD²

⇒ AD² = AB² – BD²

⇒ AD² = (15)² – (9)²

⇒ AD² = 225 – 81

⇒ AD² = 144

⇒ AD – 12 cm

(i) cos ∠ABC = Base/Hypotenuse

(In right angled ∆ABD, ∠ABC = ∠ABD)

= BD / AB

= 9/15

= 3/5

(ii) sin ∠ACB = sin ∠ACD

= perpendicular/ Hypotenuse

= AD/AC

= 12/15

= 4/5

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