# The sides of a triangle $ABC$ are $a,b$ and $c$ with $A,B$ and $C$as their opposite vertices respectively.

If $2b = 3a$ and ${\tan ^2}A = \dfrac{3}{5},$prove that there are two values of the third side, one of which is double the other.

Last updated date: 20th Mar 2023

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Answer

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Hint: Using the equation ${\tan ^2}A = \dfrac{3}{5},$find the value of $\cos A$ and then apply cosine rule.

From the question:

${\tan ^2}A = \dfrac{3}{5}$and we know that $1 + {\tan ^2}A = {\sec ^2}A$,

$

\Rightarrow {\sec ^2}A = 1 + \dfrac{3}{5}, \\

\Rightarrow {\sec ^2}A = \dfrac{8}{5}. \\

$

Further, we know that ${\cos ^2}A = \dfrac{1}{{{{\sec }^2}A}},$

$

\Rightarrow {\cos ^2}A = \dfrac{5}{8}, \\

\Rightarrow \cos A = \sqrt {\dfrac{5}{8}} , \\

\Rightarrow \cos A = \dfrac{{\sqrt 5 }}{{2\sqrt 2 }}. \\

$

Now, applying cosine rule,

$ \Rightarrow {b^2} + {c^2} - {a^2} = 2bc\cos A.$

From question, we know that $2b = 3a.$ Putting $a = \dfrac{{2b}}{3}$ and $\cos A = \dfrac{{\sqrt 5 }}{{2\sqrt 2 }}$in above equation:

$

\Rightarrow {b^2} + {c^2} - \dfrac{{4{b^2}}}{9} = 2bc \times \dfrac{{\sqrt 5 }}{{2\sqrt 2 }}, \\

\Rightarrow {c^2} + \dfrac{{5{b^2}}}{9} = \sqrt {\dfrac{5}{2}} bc, \\

\Rightarrow {c^2} - \sqrt {\dfrac{5}{2}} bc + \dfrac{{5{b^2}}}{9} = 0. \\

$

Above equation is a quadratic equation in $c.$ Let ${c_1},{c_2}$ be the root of the equation. Then,

Sum of roots $ = {c_1} + {c_2} = \sqrt {\dfrac{5}{2}} b .....(i)$

Product of root $ = {c_1}{c_2} = \dfrac{{5{b^2}}}{9}.$

We know that, ${({c_1} - {c_2})^2} = {({c_1} + {c_2})^2} - 4{c_1}{c_2}.$Putting values from above:

$

\Rightarrow {({c_1} - {c_2})^2} = \dfrac{{5{b^2}}}{2} - 4 \times \dfrac{{5{b^2}}}{9}, \\

\Rightarrow {({c_1} - {c_2})^2} = \dfrac{{5{b^2}}}{{18}}, \\

\Rightarrow {c_1} - {c_2} = \dfrac{1}{3}\sqrt {\dfrac{5}{2}} b .....(ii) \\

$

Now adding equation $(i)$and $(ii)$, we have:

$

\Rightarrow 2{c_1} = \sqrt {\dfrac{5}{2}} b + \dfrac{1}{3}\sqrt {\dfrac{5}{2}} b, \\

\Rightarrow 2{c_1} = \dfrac{{4b}}{3}\sqrt {\dfrac{5}{2}} , \\

\Rightarrow {c_1} = \dfrac{{2b}}{3}\sqrt {\dfrac{5}{2}} . \\

$

Putting the value of ${c_1}$in equation $(i)$,we’ll get:

$

\Rightarrow {c_1} + {c_2} = \sqrt {\dfrac{5}{2}} b, \\

\Rightarrow {c_2} = \sqrt {\dfrac{5}{2}} b - {c_1}, \\

\Rightarrow {c_2} = \sqrt {\dfrac{5}{2}} b - \dfrac{{2b}}{3}\sqrt {\dfrac{5}{2}} , \\

\Rightarrow {c_2} = \dfrac{b}{3}\sqrt {\dfrac{5}{2}} . \\

$

So, we have ${c_1} = \dfrac{{2b}}{3}\sqrt {\dfrac{5}{2}} $and ${c_2} = \dfrac{b}{3}\sqrt {\dfrac{5}{2}} $. From this we can say that ${c_1} = 2{c_2}.$

Therefore, we have two values of the third side in which one is twice the other. This is the required proof.

Note: Whenever two sides of a triangle are given along with the angle made by these two sides, we can use cosine rule to find out the third side.

From the question:

${\tan ^2}A = \dfrac{3}{5}$and we know that $1 + {\tan ^2}A = {\sec ^2}A$,

$

\Rightarrow {\sec ^2}A = 1 + \dfrac{3}{5}, \\

\Rightarrow {\sec ^2}A = \dfrac{8}{5}. \\

$

Further, we know that ${\cos ^2}A = \dfrac{1}{{{{\sec }^2}A}},$

$

\Rightarrow {\cos ^2}A = \dfrac{5}{8}, \\

\Rightarrow \cos A = \sqrt {\dfrac{5}{8}} , \\

\Rightarrow \cos A = \dfrac{{\sqrt 5 }}{{2\sqrt 2 }}. \\

$

Now, applying cosine rule,

$ \Rightarrow {b^2} + {c^2} - {a^2} = 2bc\cos A.$

From question, we know that $2b = 3a.$ Putting $a = \dfrac{{2b}}{3}$ and $\cos A = \dfrac{{\sqrt 5 }}{{2\sqrt 2 }}$in above equation:

$

\Rightarrow {b^2} + {c^2} - \dfrac{{4{b^2}}}{9} = 2bc \times \dfrac{{\sqrt 5 }}{{2\sqrt 2 }}, \\

\Rightarrow {c^2} + \dfrac{{5{b^2}}}{9} = \sqrt {\dfrac{5}{2}} bc, \\

\Rightarrow {c^2} - \sqrt {\dfrac{5}{2}} bc + \dfrac{{5{b^2}}}{9} = 0. \\

$

Above equation is a quadratic equation in $c.$ Let ${c_1},{c_2}$ be the root of the equation. Then,

Sum of roots $ = {c_1} + {c_2} = \sqrt {\dfrac{5}{2}} b .....(i)$

Product of root $ = {c_1}{c_2} = \dfrac{{5{b^2}}}{9}.$

We know that, ${({c_1} - {c_2})^2} = {({c_1} + {c_2})^2} - 4{c_1}{c_2}.$Putting values from above:

$

\Rightarrow {({c_1} - {c_2})^2} = \dfrac{{5{b^2}}}{2} - 4 \times \dfrac{{5{b^2}}}{9}, \\

\Rightarrow {({c_1} - {c_2})^2} = \dfrac{{5{b^2}}}{{18}}, \\

\Rightarrow {c_1} - {c_2} = \dfrac{1}{3}\sqrt {\dfrac{5}{2}} b .....(ii) \\

$

Now adding equation $(i)$and $(ii)$, we have:

$

\Rightarrow 2{c_1} = \sqrt {\dfrac{5}{2}} b + \dfrac{1}{3}\sqrt {\dfrac{5}{2}} b, \\

\Rightarrow 2{c_1} = \dfrac{{4b}}{3}\sqrt {\dfrac{5}{2}} , \\

\Rightarrow {c_1} = \dfrac{{2b}}{3}\sqrt {\dfrac{5}{2}} . \\

$

Putting the value of ${c_1}$in equation $(i)$,we’ll get:

$

\Rightarrow {c_1} + {c_2} = \sqrt {\dfrac{5}{2}} b, \\

\Rightarrow {c_2} = \sqrt {\dfrac{5}{2}} b - {c_1}, \\

\Rightarrow {c_2} = \sqrt {\dfrac{5}{2}} b - \dfrac{{2b}}{3}\sqrt {\dfrac{5}{2}} , \\

\Rightarrow {c_2} = \dfrac{b}{3}\sqrt {\dfrac{5}{2}} . \\

$

So, we have ${c_1} = \dfrac{{2b}}{3}\sqrt {\dfrac{5}{2}} $and ${c_2} = \dfrac{b}{3}\sqrt {\dfrac{5}{2}} $. From this we can say that ${c_1} = 2{c_2}.$

Therefore, we have two values of the third side in which one is twice the other. This is the required proof.

Note: Whenever two sides of a triangle are given along with the angle made by these two sides, we can use cosine rule to find out the third side.

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