Question

In \[\vartriangle {\text{ABC}}\]\[{\text{a,b,A}}\] are given and \[{{\text{c}}_1}{\text{,}}{{\text{c}}_2}\] are two values of the third side\[{\text{c}}\]. The sum of the areas of the two triangles with sides \[{\text{a,b,}}{{\text{c}}_1}\] and \[{\text{a,b,}}{{\text{c}}_2}\]is
A. \[\left( {\dfrac{{\text{1}}}{{\text{2}}}} \right){{\text{b}}^{\text{2}}}{\text{sin2A}}\]
B. \[\left( {\dfrac{{\text{1}}}{{\text{2}}}} \right){{\text{a}}^{\text{2}}}{\text{sin2A}}\]
C. \[{{\text{b}}^{\text{2}}}{\text{sin2A}}\]
D. None of these

Answer 1

Hint: In this sum we are going to see some trigonometric identities.
We have to find the sum of the areas of the two triangles using trigonometric identities.
We will use the below mentioned identities to get our answer.
\[{\text{sin2A}}\]\[{\text{ = }}\]\[{\text{2sinAcosA}}\]
\[{\text{cosA = }}\dfrac{{{{\text{b}}^{\text{2}}}{\text{ + }}{{\text{c}}^{\text{2}}}{\text{ - }}{{\text{a}}^{\text{2}}}}}{{{\text{2bc}}}}\]
\[\dfrac{{\text{a}}}{{{\text{sinA}}}}{\text{ = }}\dfrac{{\text{b}}}{{{\text{sinB}}}}\]

Complete step-by-step answer:
It is given \[{\text{a,b,A}}\] and \[{{\text{c}}_1}{\text{,}}{{\text{c}}_2}\] are two values of the third side\[{\text{c}}\].
From cosine rule,
\[{\text{cosA = }}\dfrac{{{{\text{b}}^{\text{2}}}{\text{ + }}{{\text{c}}^{\text{2}}}{\text{ - }}{{\text{a}}^{\text{2}}}}}{{{\text{2bc}}}}\]
By Cross-Multiplying we get that,
$\Rightarrow$\[{\text{(2bc)(cosA) = }}{{\text{b}}^{\text{2}}}{\text{ + }}{{\text{c}}^{\text{2}}}{\text{ - }}{{\text{a}}^{\text{2}}}\]
Bringing the LHS terms to RHS we get,
$\Rightarrow$\[{{\text{c}}^2}{\text{ - 2bc cosA + (}}{{\text{b}}^2}{\text{ - }}{{\text{a}}^2}{\text{) = 0}}\]
Let \[{{\text{c}}_1}{\text{,}}{{\text{c}}_2}\] be the roots of the above equation
Then,
The sum of roots \[{\text{ = }}{{\text{c}}_{\text{1}}}{\text{ + }}{{\text{c}}_{\text{2}}}\]\[{\text{ = 2b cosA}}\]
Multiplying \[{\text{2Rsin}}\]on the above sum of roots, we get that,
$\Rightarrow$\[{\text{2Rsin}}{{\text{C}}_{\text{1}}}{\text{ + 2Rsin}}{{\text{C}}_{\text{2}}}{\text{ = 2cosA(2RsinB)}}\]
On LHS, taking out the common term 2R we get,
$\Rightarrow$\[{\text{2R(sin}}{{\text{C}}_{\text{1}}}{\text{ + sin}}{{\text{C}}_{\text{2}}}{\text{) = 2cosA (2RsinB)}}\]
By cancelling 2R we have,
$\Rightarrow$\[{\text{sin}}{{\text{C}}_{\text{1}}}{\text{ + sin}}{{\text{C}}_{\text{2}}}{\text{ = 2cosA sinB}}\]\[...........................{\text{(1)}}\]
Here, we are using the formula of area of triangle,
Area of triangle = \[\dfrac{{\text{1}}}{{\text{2}}}{{ \times b \times h}}\]
So that,
The sum of area of two triangles is, \[\Delta {\text{ = }}{\Delta _1}{\text{ + }}{\Delta _2}\]
The sum of the areas of the two triangles with sides \[{\text{a,b,}}{{\text{c}}_1}\]and \[{\text{a,b,}}{{\text{c}}_2}\] is
$\Rightarrow$\[\Delta \] \[{\text{ = }}\]\[\dfrac{{\text{1}}}{{\text{2}}}{\text{ab sin}}{{\text{C}}_{\text{1}}}{\text{ + }}\dfrac{{\text{1}}}{{\text{2}}}{\text{ab sin}}{{\text{C}}_{\text{2}}}\]
Taking out \[\dfrac{{\text{1}}}{{\text{2}}}{\text{ab}}\] which is the common term, we have
\[\dfrac{{\text{1}}}{{\text{2}}}{\text{ab(sin}}{{\text{C}}_{\text{1}}}{\text{ + sin}}{{\text{C}}_{\text{2}}}{\text{)}}\]
From equation (1) we get,
\[{\text{ = }}\]\[\dfrac{{\text{1}}}{{\text{2}}}{\text{ab(2cosAsinB)}}\]
After Cancelling the terms, we get,
\[{\text{ = }}\]\[{\text{ab cosAsinB}}\] ……………….. (2)
[Since, we know that the formula,
\[\dfrac{{\text{a}}}{{{\text{sinA}}}}{\text{ = }}\dfrac{{\text{b}}}{{{\text{sinB}}}}\]
After cross multiplying, we get,
$\Rightarrow$\[{\text{asinB = bsinA}}\]]
Substitute \[{\text{asinB = bsinA}}\] in the equation (2) we have that
\[{\text{ = }}\]\[{\text{bsinA(b cosA)}}\]
Here, we are multiplying them. Now we have,
\[{\text{ = }}\]\[{{\text{b}}^{\text{2}}}{\text{ sinAcosA}}\]
Multiplying and dividing them by 2, we get,
\[{\text{ = }}\]\[\dfrac{{{{\text{b}}^{\text{2}}}{\text{ 2sinAcosA}}}}{2}\]
We know that the formula, \[{\text{sin2A}}\]\[{\text{ = }}\]\[{\text{2sinAcosA}}\]
Substitute above formula on them we get,
\[\Delta \]\[{\text{ = }}\]\[\dfrac{{{{\text{b}}^{\text{2}}}{\text{ sin2A}}}}{2}\]

The sum of the areas of the two triangles with sides \[{\text{a,b,}}{{\text{c}}_1}\] and \[{\text{a,b,}}{{\text{c}}_2}\] is \[\Delta \]\[{\text{ = }}\]\[\dfrac{{{{\text{b}}^{\text{2}}}{\text{ sin2A}}}}{2}\]

Note: In this question we have alternative method as follows:
From cosine rule,
\[{\text{cosA = }}\dfrac{{{{\text{b}}^{\text{2}}}{\text{ + }}{{\text{c}}^{\text{2}}}{\text{ - }}{{\text{a}}^{\text{2}}}}}{{{\text{2bc}}}}\]
By Cross-Multiplying we get that,
$\Rightarrow$\[{\text{(2bc)(cosA) = }}{{\text{b}}^{\text{2}}}{\text{ + }}{{\text{c}}^{\text{2}}}{\text{ - }}{{\text{a}}^{\text{2}}}\]
Bringing the LHS terms to RHS we get,
$\Rightarrow$\[{{\text{c}}^2}{\text{ - 2bc cosA + (}}{{\text{b}}^2}{\text{ - }}{{\text{a}}^2}{\text{) = 0}}\]
Let \[{{\text{c}}_1}{\text{,}}{{\text{c}}_2}\] be the roots of the above equation,
The sum of roots \[{\text{ = }}{{\text{c}}_{\text{1}}}{\text{ + }}{{\text{c}}_{\text{2}}}\]\[{\text{ = 2b cosA}}\]
Now, The sum of area of two triangle, \[\Delta {\text{ = }}{\Delta _1}{\text{ + }}{\Delta _2}\]
$\Rightarrow$\[{\Delta _1}\]\[{\text{ = }}\dfrac{{{\text{b}}{{\text{C}}_{\text{1}}}{\text{sinA}}}}{2}\], \[{\Delta _2}{\text{ = }}\dfrac{{{\text{b}}{{\text{C}}_2}{\text{sinA}}}}{2}\]
\[\Delta {\text{ = }}{\Delta _1}{\text{ + }}{\Delta _2}\]
\[\Delta \]\[{\text{ = }}\dfrac{{{\text{b}}{{\text{C}}_{\text{1}}}{\text{sinA + b}}{{\text{C}}_{\text{2}}}{\text{sinA}}}}{{\text{2}}}\]
Taking out the common terms in the following,
$\Rightarrow$\[\Delta \]\[{\text{ = }}\dfrac{{{\text{b(}}{{\text{C}}_{\text{1}}}{\text{ + }}{{\text{C}}_{\text{2}}}{\text{)sinA}}}}{{\text{2}}}\]
We know\[{{\text{C}}_1}{\text{ + }}{{\text{C}}_2}\]\[{\text{ = 2b cosA}}\] substitute that in the above term,
$\Rightarrow$\[\Delta \]\[{\text{ = }}\dfrac{{{\text{b(2bcosA)sinA}}}}{{\text{2}}}\]
After cancelling them, we get,
$\Rightarrow$\[\Delta \]\[{\text{ = }}\]\[{{\text{b}}^{\text{2}}}{\text{ sinAcosA}}\]
Multiplying and dividing the above terms by 2, we have that
$\Rightarrow$\[\Delta = \dfrac{{{{\text{b}}^{\text{2}}}{\text{ 2sinAcosA}}}}{2}\]
We already know that \[{\text{2sinAcosA}}\]\[{\text{ = }}\] \[{\text{sin2A}}\]
So, By Substituting those on below we get,
$\Rightarrow$\[\Delta \]\[{\text{ = }}\]\[\dfrac{{{{\text{b}}^{\text{2}}}{\text{ sin2A}}}}{2}\]
The sum of the areas of the two triangles with sides \[{\text{a,b,}}{{\text{c}}_1}\] and \[{\text{a,b,}}{{\text{c}}_2}\] is \[\Delta \]\[{\text{ = }}\]\[\dfrac{{{{\text{b}}^{\text{2}}}{\text{ sin2A}}}}{2}\]