Question

Consider $\Delta$ACB, right-angled at C, in which $AB=29units,BC=21units$ and $\angle ABC=\theta$. Determine the value of $si{n}^2\theta +co{s}^2\theta$

Answer 1

In right angled triangle $ABC$,

${\text{hypotenuse}}^2={\text{perpendicular}}^2+{\text{base}}^2$

${AB}^2={AC}^2+{BC}^2$

$\Rightarrow {AC}^2={AB}^2-{BC}^2$

$\Rightarrow {AC}^2={\left(29\right)}^2-{\left(21\right)}^2$

$\Rightarrow {AC}^2=841-441$

$\Rightarrow AC=\sqrt{400}=20$

Therefore,

$\sin \theta =\frac{\text{Perpendicular}}{\text{Hypotenuse}}$

$\Rightarrow \sin \theta =\frac{AC}{AB}$

$\Rightarrow \sin \theta =\frac{20}{29}$

Similarly,

$\cos \theta =\frac{\text{Base}}{\text{Hypotenuse}}$

$\Rightarrow \cos \theta =\frac{BC}{AB}$

$\Rightarrow \cos \theta =\frac{21}{29}$

$\therefore {\sin }^2\theta +{\cos }^2\theta$ $={\left(\frac{20}{29}\right)}^2+{\left(\frac{20}{29}\right)}^2=\frac{400}{841}+\frac{441}{841}=\frac{400+441}{841}=1$

Hence the value of ${\sin }^2\theta +{\cos }^2\theta$ is $1$.