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
Answers
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
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