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If in a ΔABC, cos B = (sin A)/(2 sin C), then the triangle is
- a)isosceles triangle
- b)equilateral triangle
- c)right angled triangle
- d)scalene triangle
Correct answer is option 'A'. Can you explain this answer?
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If in a ΔABC, cos B = (sin A)/(2 sin C), then the triangle isa)is...
Given: Cos B = (Sin A)/(2 Sin C)
To prove: ABC is an Isosceles Triangle
Proof:
Let us draw a triangle ABC.
From the given equation, we have:
Cos B = (Sin A)/(2 Sin C)
We know that:
Cos B = (AC/AB)
Sin A = (BC/AB)
Sin C = (AC/BC)
Substituting the above values in the given equation, we get:
AC/AB = (BC/AB)/(2(AC/BC))
=> AC/AB = BC/2AC
=> AC^2 = AB × BC/2
Now, let us apply the Law of Cosines in Triangle ABC.
According to Law of Cosines:
BC^2 = AB^2 + AC^2 - 2AB × AC × Cos B
Substituting the value of AC^2 from the previous equation, we get:
BC^2 = AB^2 + AB × BC/2 - AB × BC × Cos B
=> BC^2 = AB^2 + AB × BC/2 - AB × (BC^2 + AB^2 - BC^2)/2AB
Simplifying the above equation, we get:
BC^2 = AB^2/4
=> BC = AB/2
Hence, we have proved that AB = BC. Therefore, ABC is an Isosceles Triangle.
To prove: ABC is an Isosceles Triangle
Proof:
Let us draw a triangle ABC.
From the given equation, we have:
Cos B = (Sin A)/(2 Sin C)
We know that:
Cos B = (AC/AB)
Sin A = (BC/AB)
Sin C = (AC/BC)
Substituting the above values in the given equation, we get:
AC/AB = (BC/AB)/(2(AC/BC))
=> AC/AB = BC/2AC
=> AC^2 = AB × BC/2
Now, let us apply the Law of Cosines in Triangle ABC.
According to Law of Cosines:
BC^2 = AB^2 + AC^2 - 2AB × AC × Cos B
Substituting the value of AC^2 from the previous equation, we get:
BC^2 = AB^2 + AB × BC/2 - AB × BC × Cos B
=> BC^2 = AB^2 + AB × BC/2 - AB × (BC^2 + AB^2 - BC^2)/2AB
Simplifying the above equation, we get:
BC^2 = AB^2/4
=> BC = AB/2
Hence, we have proved that AB = BC. Therefore, ABC is an Isosceles Triangle.
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If in a ΔABC, cos B = (sin A)/(2 sin C), then the triangle isa)isosceles triangleb)equilateral trianglec)right angled triangled)scalene triangleCorrect answer is option 'A'. Can you explain this answer?
Question Description
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