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Question: If in a triangle ABC, 2 cos A = sin B cosec C, then (a) a = h (b) b = c (c) c = a (d) 2a = ...

If in a triangle ABC, 2 cos A = sin B cosec C, then
(a) a = h
(b) b = c
(c) c = a
(d) 2a = bc

Explanation

Solution

Hint: To solve this question given above, we will first find out the value of sin B from the relation given in the question. Then we will make use of the fact that the sum of all the angles in a triangle is 180o.{{180}^{o}}. From there, we will find the value of B in terms of A and C and then we will put this value in sin B. Now, we will write these trigonometric ratios in terms of the side length.

Complete step-by-step answer:
To start with, we are given in the question that,
2cos⁡A=sin⁡Bcosec⁡C.....(i)2\cos A=\sin B\operatorname{cosec}C.....\left( i \right)
Now, we know that, cosec⁡θ=1sin⁡θ.\operatorname{cosec}\theta =\dfrac{1}{\sin \theta }. Thus, applying this formula in (i), we will get,
⇒2cos⁡A=sin⁡Bsin⁡C\Rightarrow 2\cos A=\dfrac{\sin B}{\sin C}
⇒2cos⁡Asin⁡C=sin⁡B\Rightarrow 2\cos A\sin C=\sin B
⇒sin⁡B=2cos⁡Asin⁡C.....(ii)\Rightarrow \sin B=2\cos A\sin C.....\left( ii \right)
Now, we will draw a triangle ABC.

Now, we know that the sum of all the angles in the triangle is 180o.{{180}^{o}}. Thus, we will get,
A+B+C=180oA+B+C={{180}^{o}}
⇒A+C=180o−B\Rightarrow A+C={{180}^{o}}-B
⇒B=180o−(A+C).....(iii)\Rightarrow B={{180}^{o}}-\left( A+C \right).....\left( iii \right)
Now, we will apply sine on both sides. Thus, we will get,
⇒sin⁡B=sin⁡[180o−(A+C)]\Rightarrow \sin B=\sin \left[ {{180}^{o}}-\left( A+C \right) \right]
We know that, sin⁡(180o−θ)=sin⁡θ,\sin \left( {{180}^{o}}-\theta \right)=\sin \theta , so we will get,
⇒sin⁡B=sin⁡(A+C)....(iv)\Rightarrow \sin B=\sin \left( A+C \right)....\left( iv \right)
From (ii) and (iv), we will get,
⇒sin⁡(A+C)=2cos⁡Asin⁡C\Rightarrow \sin \left( A+C \right)=2\cos A\sin C
Now, we will apply the following identity in the above equation,
sin⁡(x+y)=sin⁡xcos⁡y+cos⁡xsin⁡y\sin \left( x+y \right)=\sin x\cos y+\cos x\sin y
Thus, we will get,
⇒sin⁡Acos⁡C+cos⁡Asin⁡C=2cos⁡Asin⁡C\Rightarrow \sin A\cos C+\cos A\sin C=2\cos A\sin C
⇒sin⁡Acos⁡C=cos⁡Asin⁡C.....(v)\Rightarrow \sin A\cos C=\cos A\sin C.....\left( v \right)
Now, we know that,
sin⁡Aa=sin⁡Cc=k\dfrac{\sin A}{a}=\dfrac{\sin C}{c}=k
Also,
cos⁡A=b2+c2−a22bc\cos A=\dfrac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}
cos⁡C=a2+b2−c22ab\cos C=\dfrac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ab}
Thus, we will put these values in equation (v). After doing this, we will get,
⇒ka(a2+b2−c22ab)=(b2+c2−a22bc)kc\Rightarrow ka\left( \dfrac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ab} \right)=\left( \dfrac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc} \right)kc
⇒a(a2+b2−c22ab)=c(b2+c2−a22bc)\Rightarrow a\left( \dfrac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ab} \right)=c\left( \dfrac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc} \right)
⇒a2+b2−c22b=b2+c2−a22b\Rightarrow \dfrac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2b}=\dfrac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2b}
⇒a2+b2−c2=b2+c2−a2\Rightarrow {{a}^{2}}+{{b}^{2}}-{{c}^{2}}={{b}^{2}}+{{c}^{2}}-{{a}^{2}}
⇒2a2=2c2\Rightarrow 2{{a}^{2}}=2{{c}^{2}}
⇒a2=c2\Rightarrow {{a}^{2}}={{c}^{2}}
⇒a=c\Rightarrow a=c
Hence, the option (c) is the right answer.

Note: The above question can also be solved by an alternate method as shown below:
2cos⁡A=sin⁡Bcosec⁡C2\cos A=\sin B\operatorname{cosec}C
Now, we will write cosec C in terms of sin C with the help of the following relation:
cosec⁡C=1sin⁡C\operatorname{cosec}C=\dfrac{1}{\sin C}
Thus, we will get the following equation:
2cos⁡A=sin⁡Bsin⁡C2\cos A=\dfrac{\sin B}{\sin C}
Now, we know that,
sin⁡Bb=sin⁡Cc\dfrac{\sin B}{b}=\dfrac{\sin C}{c}
⇒sin⁡Bsin⁡C=bc\Rightarrow \dfrac{\sin B}{\sin C}=\dfrac{b}{c}
Thus, we will get the following equation,
⇒2cos⁡A=bc\Rightarrow 2\cos A=\dfrac{b}{c}
Now, we will use the formula of cos A in the above equation which is shown below:
cos⁡A=b2+c2−a22bc\cos A=\dfrac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}
Thus, we will get,
⇒2(b2+c2−a22bc)=bc\Rightarrow 2\left( \dfrac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc} \right)=\dfrac{b}{c}
⇒b2+c2−a2bc=bc\Rightarrow \dfrac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{bc}=\dfrac{b}{c}
⇒b2+c2−a2=b2\Rightarrow {{b}^{2}}+{{c}^{2}}-{{a}^{2}}={{b}^{2}}
⇒c2−a2=0\Rightarrow {{c}^{2}}-{{a}^{2}}=0
⇒c2=a2\Rightarrow {{c}^{2}}={{a}^{2}}
⇒c=a\Rightarrow c=a