**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,

AB^{2 }= AD^{2 }+ BD^{2}

⇒ AD^{2 }= AB^{2} – BD^{2}

⇒ AD^{2} = (15)^{2} – (9)^{2}

⇒ AD^{2} = 225 – 81

⇒ AD^{2 }= 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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