In A ABC, AB= AC = 15 cm. , BC = 18 cm. find cos < ABC.
2
(a)2/5
(b) 4/5(c)1/5 (d)3/5
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2 thoughts on “In A ABC, AB= AC = 15 cm. , BC = 18 cm. find cos < ABC.<br />2<br />(a)2/5<br />(b) 4/5(c)1/5 (d)3/5<br />​”

1. 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,

   By Pythagoras theorem, we get

   AB^2 = AD^2 + BD^2

   putting the respective values, we get

   AD = 12 cm

   (i) cos ∠ ABC = Base/ Hypotenuse

   (In right angled ΔABD,∠ABC=∠ABD)

   = BD / AB

   = 9/15

   = 3/5

   (ii) sin ∠ACB=sin∠ACD

   = perpendicular/ Hypotenuse

   = AD/AC

   = 12/15

   = 4/5

   Step-by-step explanation:

   thank❤ you

   Reply

2. Answer:

   option d

   Step-by-step explanation:

   AB = AC = 15 cm

   BC = 18 cm

   it is an isosceles triangle

   if we draw AD ⊥ from A to line BC

   it will bisect BC so BD = 18/2 = 9 cm

   Δ ABD will be right angle triangle

   with Base BD – 9 cm & Hypotenuse = AB = 15 cm

   ∠ABD = ∠ABC ( as D is a point on line BC)

   Cos∠ABC = 9/15

   => Cos∠ABC = 3/5

   => Cos∠ABC = 0.6

   Reply