Question

In an isosceles ΔABC, AB = AC. If ∠A = 20°, then


a = 2b sin10°


a = 4b sin10°


a3 + b3 = 3ab2


a3 + b3 = 3ac2

Answer 1

∵∠A=20

0

∴∠B=∠C=80

0

⇒△ABC is isosceles.

⇒AB=AC

⇒c=b

Using sine rule,

∴

sin20

a

=

sin80

b

=

sin80

c

Rightarrow

sin20

a

=

sin(90−10)

b

=

sin80

c

⇒

sin20

a

=

cos10

b

=

cos10

c

⇒a=

cos10

bsin20

=

cos10

b.2sin10cos10

=2bsin10

0

∴a

3

+b

3

=(2bsin10)

3

+b

3

=8b

3

sin

3

10+b

3

=b

3

(8sin

3

10+1)

=b

3

(2(4sin

3

10+1))

Using the trigonometric formula sin3A=3sinA−4sin

3

A or 4sin

3

A=3sinA−sin3Aabove we get

=b

3

(2(3sin10−sin30)+1)

=b

3

(6sin10−2×

2

1

+1)

=b

3

(6sin10)

=3b

2

(2bsin10)

We know that a=2bsin10

∴a

3

+b

3

=3b

2

(2bsin10)(from above)

=3b

2

a=3c 2a (as b=c)