in an isosceles triangle ABC AB=AC angleA =3 angleB, then angle A
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Answer:
Here's the correct answer
Step-by-step explanation:
∵∠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
(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
2
a (as b=c)
Hope you can understand..
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