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

In the given figure, AD=DBAD = DB and B\angle B is a right angle. Determine:
sin2θ+cos2θ{\sin ^2}\theta + {\cos ^2}\theta

Explanation

Solution

Use the given data AD=DBAD = DB to find the length of DBDB and apply the Pythagoras theorem in the triangle ABC to find BCBC. 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=DBAD = DB and B\angle B is a right angle.
We can use the given data,
AB=aAB = a
ABAB can be break in two parts as ADAD and DBDB, so it can be express as:
AD+DB=aAD + DB = a
It is also given that AD=DBAD = DB, so we have from the above equation:
AD+AD=aAD + AD = a
2AD=a2AD = a
AD=a2AD = \dfrac{a}{2}
Thus, we have the conclusion that:
AD=DB=a2AD = DB = \dfrac{a}{2}.
Now, apply the Pythagoras theorem in the triangle ABCABC, then we have
AC2=AB2+BC2A{C^2} = A{B^2} + B{C^2}
Substitute the value of AB=aAB = a and AC=bAC = b into the equation, then we obtain
b2=a2+BC2{b^2} = {a^2} + B{C^2}
Solve the equation for the value of BCBC,
BC2=b2a2\Rightarrow B{C^2} = {b^2} - {a^2}
BC=b2a2\Rightarrow BC = \sqrt {{b^2} - {a^2}}
Now, we have in the ΔBCD\Delta BCD:
Base (BC)=b2a2\left( {BC} \right) = \sqrt {{b^2} - {a^2}} and Perpendicular(BD)=a2\left( {BD} \right) = \dfrac{a}{2}
Now, apply the Pythagoras theorem in ΔBCD\Delta BCD, so we have
BC2+BD2=CD2B{C^2} + B{D^2} = C{D^2}
Substitute the value of BCBC and BDBD into the equation:
(b2a2)2+(a2)2=CD2{\left( {\sqrt {{b^2} - {a^2}} } \right)^2} + {\left( {\dfrac{a}{2}} \right)^2} = C{D^2}
CD2=b2a2+a24\Rightarrow C{D^2} = {b^2} - {a^2} + \dfrac{{{a^2}}}{4}
Simplify the equation:
CD2=4b24a2+a24\Rightarrow C{D^2} = \dfrac{{4{b^2} - 4{a^2} + {a^2}}}{4}
CD2=4b23a24\Rightarrow C{D^2} = \dfrac{{4{b^2} - 3{a^2}}}{4}
CD=4b23a22\Rightarrow CD = \dfrac{{\sqrt {4{b^2} - 3{a^2}} }}{2}
Now, we have in the ΔBCD\Delta BCD:
Base (BC)=b2a2\left( {BC} \right) = \sqrt {{b^2} - {a^2}} , Perpendicular (BD)=a2\left( {BD} \right) = \dfrac{a}{2} and the hypotenuse (CD)=4b23a22\left( {CD} \right) = \dfrac{{\sqrt {4{b^2} - 3{a^2}} }}{2}
Now, use the trigonometric ratio in ΔBCD\Delta BCD,
sinθ=BDCD\sin \theta = \dfrac{{BD}}{{CD}}
Substitute the values of BDBD and CDCD, so we have
sinθ=a24b23a22\sin \theta = \dfrac{{\dfrac{a}{2}}}{{\dfrac{{\sqrt {4{b^2} - 3{a^2}} }}{2}}}
sinθ=a4b23a2\sin \theta = \dfrac{a}{{\sqrt {4{b^2} - 3{a^2}} }}
Using the trigonometric ratio:
cosθ=BCCD\cos \theta = \dfrac{{BC}}{{CD}}
Substitute the values of BCBC and CDCD, so we have
cosθ=b2a24b23a22\cos \theta = \dfrac{{\sqrt {{b^2} - {a^2}} }}{{\dfrac{{\sqrt {4{b^2} - 3{a^2}} }}{2}}}
cosθ=2b2a24b23a2\cos \theta = \dfrac{{2\sqrt {{b^2} - {a^2}} }}{{\sqrt {4{b^2} - 3{a^2}} }}
We have to find the value of sin2θ+cos2θ{\sin ^2}\theta + {\cos ^2}\theta , so substitute the value of sinθ\sin \theta and cosθ\cos \theta into the equation:
sin2θ+cos2θ=(a4b23a2)2+(2b2a24b23a2)2{\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}
sin2θ+cos2θ=a24b23a2+4(b2a2)4b23a2\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}}}
sin2θ+cos2θ=a2+4b24a24b23a2\Rightarrow {\sin ^2}\theta + {\cos ^2}\theta = \dfrac{{{a^2} + 4{b^2} - 4{a^2}}}{{4{b^2} - 3{a^2}}}
sin2θ+cos2θ=4b23a24b23a2\Rightarrow {\sin ^2}\theta + {\cos ^2}\theta = \dfrac{{4{b^2} - 3{a^2}}}{{4{b^2} - 3{a^2}}}
sin2θ+cos2θ=1\Rightarrow {\sin ^2}\theta + {\cos ^2}\theta = 1
Therefore, the value of sin2θ+cos2θ{\sin ^2}\theta + {\cos ^2}\theta is 11.

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.