Solveeit Logo
Login

Question

Question: In \( \Delta ABC \) , if \( \sin A:\sin C = \sin (A - B):\sin (B - C) \) , then \( {a^2} \) , \( {b^...

In ΔABC\Delta ABC , if sin⁡A:sin⁡C=sin⁡(A−B):sin⁡(B−C)\sin A:\sin C = \sin (A - B):\sin (B - C) , then a2{a^2} , b2{b^2} and c2{c^2} are in
A. A.P
B. G.P
C. H.P
D. None of these

Explanation

Solution

In this question, a triangle is given which satisfy the condition sin⁡A:sin⁡C=sin⁡(A−B):sin⁡(B−C)\sin A:\sin C = \sin (A - B):\sin (B - C) , we have to find relation of a2{a^2} , b2{b^2} and c2{c^2} .
Use the property of a triangle, the sum of all angles of a triangle is 180∘{180^ \circ } , to write the angle in terms of the other two angles.
⇒\Rightarrow A+B+C=180∘A + B + C = {180^ \circ }
According to the trigonometrically ratios of angles (180∘ − θ)\left( {180^\circ {\text{ }} - {\text{ }}\theta } \right) , sin⁡(180∘ − θ)=sin⁡θ\sin \left( {180^\circ {\text{ }} - {\text{ }}\theta } \right) = \sin \theta .
Now, use the trigonometric formula, sin⁡(A+B)sin⁡(A−B)=sin⁡2A−sin⁡2B\sin (A + B)\sin (A - B) = {\sin ^2}A - {\sin ^2}B
Apply the Sine rule of the triangle,
If a, b and c are the sides of the triangle and their corresponding opposite angles are A , B and C then,
asin⁡A=bsin⁡B=csin⁡C=k\dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}} = \dfrac{c}{{\sin C}} = k
From this relation write sin⁡A\sin A , sin⁡B\sin B and sin⁡C\sin C in terms of a, b, c and k.
If a2{a^2} , b2{b^2} and c2{c^2} are in A.P. then the difference between the consecutive terms is the same.
⇒b2−a2=c2−b2\Rightarrow {b^2} - {a^2} = {c^2} - {b^2}
⇒2b2=c2+a2\Rightarrow 2{b^2} = {c^2} + {a^2}

Complete step-by-step answer:
Consider a triangle , where A, B and C are the angles and a, b, and c are the sides.
Given is the equation, sin⁡A:sin⁡C=sin⁡(A−B):sin⁡(B−C)\sin A:\sin C = \sin (A - B):\sin (B - C) which is same as,
sin⁡Asin⁡C=sin⁡(A−B)sin⁡(B−C)…(1)\dfrac{{\sin A}}{{\sin C}} = \dfrac{{\sin (A - B)}}{{\sin (B - C)}} \ldots (1)
Since the sum of all angles of a triangle is 180∘{180^ \circ } .
A+B+C=180∘A + B + C = {180^ \circ }
⇒A=180∘−(B+C)…(2)\Rightarrow A = {180^ \circ } - (B + C) \ldots (2)
And, find the angle C,
C=180∘−(A+B)…(3)C = {180^ \circ } - (A + B) \ldots (3)
Substitute the value of A from equation (1)(1) and C from equation (2)(2) into the left-hand side of the equation (3)(3) .
sin⁡Asin⁡C=sin⁡(A−B)sin⁡(B−C)\dfrac{{\sin A}}{{\sin C}} = \dfrac{{\sin (A - B)}}{{\sin (B - C)}}
sin⁡(180∘−(B+C))sin⁡(180∘−(A+B))=sin⁡(A−B)sin⁡(B−C)\dfrac{{\sin ({{180}^ \circ } - (B + C))}}{{\sin ({{180}^ \circ } - (A + B))}} = \dfrac{{\sin (A - B)}}{{\sin (B - C)}}
Apply the trigonometrically ratios of angles (180∘ − θ)\left( {180^\circ {\text{ }} - {\text{ }}\theta } \right) , sin⁡(180∘ − θ)=sin⁡θ\sin \left( {180^\circ {\text{ }} - {\text{ }}\theta } \right) = \sin \theta .
∴sin⁡(B+C)sin⁡(A+B)=sin⁡(A−B)sin⁡(B−C)\therefore \dfrac{{\sin (B + C)}}{{\sin (A + B)}} = \dfrac{{\sin (A - B)}}{{\sin (B - C)}}
After cross multiplying the equation we get,
⇒sin⁡(B+C)sin⁡(B−C)=sin⁡(A−B)sin⁡(A+B)…(4)\Rightarrow \sin (B + C)\sin (B - C) = \sin (A - B)\sin (A + B) \ldots (4)
Apply the trigonometric formula, sin⁡(A+B)sin⁡(A−B)=sin⁡2A−sin⁡2B\sin (A + B)\sin (A - B) = {\sin ^2}A - {\sin ^2}B to the equation (4)(4) .
⇒sin⁡2B−sin⁡2C=sin⁡2A−sin⁡2B\Rightarrow {\sin ^2}B - {\sin ^2}C = {\sin ^2}A - {\sin ^2}B
⇒2sin⁡2B=sin⁡2A+sin⁡2C…(5)\Rightarrow 2{\sin ^2}B = {\sin ^2}A + {\sin ^2}C \ldots (5)
Apply the Sine rule of the triangle,
If a, b and c are the sides of the triangle and their corresponding opposite angles are A , B and C then,
asin⁡A=bsin⁡B=csin⁡C=k\dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}} = \dfrac{c}{{\sin C}} = k
Take the ratio,
bsin⁡B=k\dfrac{b}{{\sin B}} = k
⇒bk=sin⁡B\Rightarrow \dfrac{b}{k} = \sin B
Now, take the ratio,
asin⁡A=k\dfrac{a}{{\sin A}} = k
⇒ak=sin⁡A\Rightarrow \dfrac{a}{k} = \sin A
Now, take the ratio,
csin⁡C=k\dfrac{c}{{\sin C}} = k
⇒ck=sin⁡C\Rightarrow \dfrac{c}{k} = \sin C
Substitute the values of sin⁡A\sin A , sin⁡B\sin B and sin⁡C\sin C into the equation (5)(5)
⇒2(bk)2=(ak)2+(ck)2\Rightarrow 2{\left( {\dfrac{b}{k}} \right)^2} = {\left( {\dfrac{a}{k}} \right)^2} + {\left( {\dfrac{c}{k}} \right)^2}
⇒2b2k2=a2+c2k2\Rightarrow \dfrac{{2{b^2}}}{{{k^2}}} = \dfrac{{{a^2} + {c^2}}}{{{k^2}}}
⇒2b2=a2+c2\Rightarrow 2{b^2} = {a^2} + {c^2}
If a2{a^2} , b2{b^2} and c2{c^2} are in A.P. then the difference between the consecutive terms is the same.
⇒b2−a2=c2−b2\Rightarrow {b^2} - {a^2} = {c^2} - {b^2}
⇒2b2=c2+a2\Rightarrow 2{b^2} = {c^2} + {a^2}

Correct Answer : A) A.P.

Note:
An arithmetic progression is a sequence of numbers such that the difference of any two successive members is a constant. That is if a, b ,c are in A.P. then, b−a=c−bb - a = c - b . A sequence of numbers is called a geometric progression if the ratio of any two consecutive terms is same i.e. For example 2,4,8,16 is a GP because ratio of any two consecutive terms in the series is same , 42 =84=168=2\dfrac{4}{2}{\text{ }} = \dfrac{8}{4} = \dfrac{{16}}{8} = 2 . A sequence of numbers is called a harmonic progression if the reciprocal of the terms are in AP. In simple terms, a ,b, c, are in HP if 1a\dfrac{1}{a} , 1b\dfrac{1}{b} , 1c\dfrac{1}{c} are in AP.