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Trigonometric Ratios — Maths STD 10 — Question

CBSE BoardEnglish MediumSTD 10MathsTrigonometric Ratios5 Marks
Question
In a $\triangle\text{ABC},\angle\text{C}=90^\circ,\angle\text{ABC}=\theta^\circ,\text{BC}=21\text{units}$ and AB = 29units.
Show that $\big(\cos^2\theta-\sin^2\theta\big)=\frac{41}{841}.$
✓

Answer


In $\triangle\text{ABC},\angle\text{C}=90^\circ$
AB = 29 units and BC = 21 units
By Pythagoras theorem, we have
$\text{AC}^2=\text{AB}^2-\text{BC}^2$
$=29^2-21^2=841-441=400$
$\Rightarrow\text{AC}=20\text{ units}$
$\therefore\ \sin\theta=\frac{\text{Perpendicular}}{\text{Hypotenuse}}=\frac{\text{AC}}{\text{AB}}=\frac{20}{29}$
$\cos\theta=\frac{\text{Base}}{\text{Hypotenuse}}=\frac{\text{BC}}{\text{AB}}=\frac{21}{29}$
$\therefore\text{L.H.S.}=\cos^2\theta-\sin^2\theta$
$=\Big(\frac{21}{29}\Big)^2-\Big(\frac{20}{29}\Big)^2$
$=\frac{441}{841}-\frac{400}{841}$
$=\frac{41}{841}$
$=\text{R.H.S.}$

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