Question
Tan θ =$\frac{21}{20}$.Find Sin θ= , Cos θ= ?
✓
Answer
$\cos \theta=\frac{21}{\sqrt{20}}$..(i) [Given]
In right angled $\triangle ABC$,
$\angle C=\theta \text {. }$
$\begin{array}{ll}
\tan \theta=\frac{\text { Opposite side of } \theta}{\text { Adjacent side of } \theta} \\
\therefore \quad \tan \theta=\frac{ AB }{ BC } \\
\therefore \quad & \frac{ AB }{ BC }=\frac{21}{20}
\end{array}$
...[From (i) and (ii)]
Let the common multiple be $k$.
$\therefore A B=21 k \text { and } B C=20 k$
Now, $A C^2=A B^2+B C^2 \ldots[$ Pythagoras theorem $]$
$=(21) K^2+(20 K)^2$
$=441 K^2-400^2$
$=841 K^2$
$\begin{aligned}
\therefore & AB =\sqrt{841 k^2} \ldots[\text { Taking square root of both sides] } \\
= & 29 K \\
& \therefore \quad \sin \theta=\frac{\text { Opposite side of } \theta}{\text { Hypotenuse }}=\frac{ AB }{ AC }=\frac{21 k }{29 k }=\frac{21}{29} \\
& \cos \theta=\frac{\text { Adjacent side of } \theta}{\text { Hypotenuse }}=\frac{ BC }{ AC }=\frac{20 k }{29 k }=\frac{ 2 0 }{ 2 9 }
\end{aligned}$
In right angled $\triangle ABC$,
$\angle C=\theta \text {. }$
$\begin{array}{ll}
\tan \theta=\frac{\text { Opposite side of } \theta}{\text { Adjacent side of } \theta} \\
\therefore \quad \tan \theta=\frac{ AB }{ BC } \\
\therefore \quad & \frac{ AB }{ BC }=\frac{21}{20}
\end{array}$
...[From (i) and (ii)]
Let the common multiple be $k$.
$\therefore A B=21 k \text { and } B C=20 k$
Now, $A C^2=A B^2+B C^2 \ldots[$ Pythagoras theorem $]$
$=(21) K^2+(20 K)^2$
$=441 K^2-400^2$
$=841 K^2$
$\begin{aligned}
\therefore & AB =\sqrt{841 k^2} \ldots[\text { Taking square root of both sides] } \\
= & 29 K \\
& \therefore \quad \sin \theta=\frac{\text { Opposite side of } \theta}{\text { Hypotenuse }}=\frac{ AB }{ AC }=\frac{21 k }{29 k }=\frac{21}{29} \\
& \cos \theta=\frac{\text { Adjacent side of } \theta}{\text { Hypotenuse }}=\frac{ BC }{ AC }=\frac{20 k }{29 k }=\frac{ 2 0 }{ 2 9 }
\end{aligned}$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
U = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
A = {2, 4, 6, 8, 10} Find A' And Draw The inference.
A = {2, 4, 6, 8, 10} Find A' And Draw The inference.
Multiply:$\sqrt{10}$ by $\sqrt{40}$
→Convert into Decimal form the rational numbers And Write is either terminating or non-terminating recurring type :$\frac{101}{8}$
→If AB = 5 cm, BP = 2 cm and AP = 3.4 cm, compare the segments.
Factorise:$5\sqrt{5}\text{x}^2-20\text{x}+3\sqrt{5}$
→Factorise:$\text{x}^2-2\sqrt{3}\text{x}-24$
→Factorize:$\text{x}^2-\sqrt{3}\text{x}-6$
→$\square \mathrm{PQRS}$ is a parallelogram. $\mathrm{PQ}=3.5, \mathrm{PS}=5.3 \angle \mathrm{Q}=50^{\circ}$ then find the lengths of remaining sides and measures of remaining angles.
→A cone and a hemisphere have equal bases and equal volumes. Find the ratio of their heights.
Find the values of : $\cos 60^{\circ} \times \cos 30^{\circ}+\sin 60^{\circ} \times \sin 30^{\circ}$
→