Question

In the given figure, $AD = DB$ and $\angle B$ is a right angle. Determine:
${\sin ^2}\theta + {\cos ^2}\theta $

Answer 1

Hint: Use the given data $AD = DB$ to find the length of $DB$ and apply the Pythagoras theorem in the triangle ABC to find $BC$. Use the length DB and BC to find the hypotenuse of the triangle BCD using Pythagoras theorem and then find the trigonometric ratios to approach the desired result.

Complete step-by-step answer:
We have given that $AD = DB$ and $\angle B$ is a right angle.
We can use the given data,
$AB = a$
$AB$ can be break in two parts as $AD$ and $DB$, so it can be express as:
$AD + DB = a$
It is also given that $AD = DB$, so we have from the above equation:
$AD + AD = a$
$2AD = a$
$AD = \dfrac{a}{2}$
Thus, we have the conclusion that:
$AD = DB = \dfrac{a}{2}$.
Now, apply the Pythagoras theorem in the triangle $ABC$, then we have
$A{C^2} = A{B^2} + B{C^2}$
Substitute the value of $AB = a$ and $AC = b$ into the equation, then we obtain
${b^2} = {a^2} + B{C^2}$
Solve the equation for the value of $BC$,
$ \Rightarrow B{C^2} = {b^2} - {a^2}$
$ \Rightarrow BC = \sqrt {{b^2} - {a^2}} $
Now, we have in the $\Delta BCD$:
Base $\left( {BC} \right) = \sqrt {{b^2} - {a^2}} $ and Perpendicular$\left( {BD} \right) = \dfrac{a}{2}$
Now, apply the Pythagoras theorem in $\Delta BCD$, so we have
$B{C^2} + B{D^2} = C{D^2}$
Substitute the value of $BC$ and $BD$ into the equation:
${\left( {\sqrt {{b^2} - {a^2}} } \right)^2} + {\left( {\dfrac{a}{2}} \right)^2} = C{D^2}$
$ \Rightarrow C{D^2} = {b^2} - {a^2} + \dfrac{{{a^2}}}{4}$
Simplify the equation:
$ \Rightarrow C{D^2} = \dfrac{{4{b^2} - 4{a^2} + {a^2}}}{4}$
\[ \Rightarrow C{D^2} = \dfrac{{4{b^2} - 3{a^2}}}{4}\]
\[ \Rightarrow CD = \dfrac{{\sqrt {4{b^2} - 3{a^2}} }}{2}\]
Now, we have in the $\Delta BCD$:
Base $\left( {BC} \right) = \sqrt {{b^2} - {a^2}} $ , Perpendicular $\left( {BD} \right) = \dfrac{a}{2}$ and the hypotenuse \[\left( {CD} \right) = \dfrac{{\sqrt {4{b^2} - 3{a^2}} }}{2}\]
Now, use the trigonometric ratio in $\Delta BCD$,
$\sin \theta = \dfrac{{BD}}{{CD}}$
Substitute the values of $BD$ and $CD$, so we have
$\sin \theta = \dfrac{{\dfrac{a}{2}}}{{\dfrac{{\sqrt {4{b^2} - 3{a^2}} }}{2}}}$
$\sin \theta = \dfrac{a}{{\sqrt {4{b^2} - 3{a^2}} }}$
Using the trigonometric ratio:
$\cos \theta = \dfrac{{BC}}{{CD}}$
Substitute the values of $BC$ and $CD$, so we have
$\cos \theta = \dfrac{{\sqrt {{b^2} - {a^2}} }}{{\dfrac{{\sqrt {4{b^2} - 3{a^2}} }}{2}}}$
$\cos \theta = \dfrac{{2\sqrt {{b^2} - {a^2}} }}{{\sqrt {4{b^2} - 3{a^2}} }}$
We have to find the value of ${\sin ^2}\theta + {\cos ^2}\theta $, so substitute the value of $\sin \theta $ and $\cos \theta $ into the equation:
\[{\sin ^2}\theta + {\cos ^2}\theta = {\left( {\dfrac{a}{{\sqrt {4{b^2} - 3{a^2}} }}} \right)^2} + {\left( {\dfrac{{2\sqrt {{b^2} - {a^2}} }}{{\sqrt {4{b^2} - 3{a^2}} }}} \right)^2}\]
\[ \Rightarrow {\sin ^2}\theta + {\cos ^2}\theta = \dfrac{{{a^2}}}{{4{b^2} - 3{a^2}}} + \dfrac{{4\left( {{b^2} - {a^2}} \right)}}{{4{b^2} - 3{a^2}}}\]
\[ \Rightarrow {\sin ^2}\theta + {\cos ^2}\theta = \dfrac{{{a^2} + 4{b^2} - 4{a^2}}}{{4{b^2} - 3{a^2}}}\]
\[ \Rightarrow {\sin ^2}\theta + {\cos ^2}\theta = \dfrac{{4{b^2} - 3{a^2}}}{{4{b^2} - 3{a^2}}}\]
\[ \Rightarrow {\sin ^2}\theta + {\cos ^2}\theta = 1\]
Therefore, the value of \[{\sin ^2}\theta + {\cos ^2}\theta \] is $1$.

Note: The Pythagoras theorem says that when one of the angles of the triangle is a right angle then the square of the hypotenuse of the triangle is equal to the sum of the squares of the perpendicular and base of the triangle.