Question
Cos θ =$\frac{35}{37}$.Find Sin θ= , Tan θ= ?
✓
Answer
$\cos \theta=\frac{35}{37} \ldots$ (i) ) [Given] In right angled $\triangle ABC$,
∠C = θ.
$\cos \theta=\frac{\text { Adjacent side of } \theta}{\text { Hypotenuse }}$
$\therefore \quad \cos \theta=\frac{ BC }{ AC }$
$\therefore \quad \frac{ BC }{ AC }=\frac{35}{37}$
Let the common multiple be k.
∴ BC = 35k and AC = 37k
Now, $AC^2 = AB^2 + BC^2 …[Pythagoras theorem]$
$\therefore (37k)^2 = AB^2+ (35k)^2$
$1369k^2 = AB^2 + 1225k^2$
$AB2 = 1369k^2 – 1225k^2$
$= 144k^2$
$AB = 144k^2$
$AB =\sqrt{2 g h K^2}$... [Taking square root of both sides]
$= 12k$
$ \therefore \quad \sin \theta =\frac{\text { Opposite side of } \theta}{\text { Hypotenuse }}=\frac{ AB }{ AC }=\frac{12 k }{37 k }=\frac{12}{37}$
$\tan \theta =\frac{\text { Opposite side of } \theta}{\text { Adjacent side of } \theta}=\frac{ AB }{ BC }=\frac{12 k }{35 k }=\frac{12}{35}$
∠C = θ.
$\cos \theta=\frac{\text { Adjacent side of } \theta}{\text { Hypotenuse }}$
$\therefore \quad \cos \theta=\frac{ BC }{ AC }$
$\therefore \quad \frac{ BC }{ AC }=\frac{35}{37}$
Let the common multiple be k.
∴ BC = 35k and AC = 37k
Now, $AC^2 = AB^2 + BC^2 …[Pythagoras theorem]$
$\therefore (37k)^2 = AB^2+ (35k)^2$
$1369k^2 = AB^2 + 1225k^2$
$AB2 = 1369k^2 – 1225k^2$
$= 144k^2$
$AB = 144k^2$
$AB =\sqrt{2 g h K^2}$... [Taking square root of both sides]
$= 12k$
$ \therefore \quad \sin \theta =\frac{\text { Opposite side of } \theta}{\text { Hypotenuse }}=\frac{ AB }{ AC }=\frac{12 k }{37 k }=\frac{12}{37}$
$\tan \theta =\frac{\text { Opposite side of } \theta}{\text { Adjacent side of } \theta}=\frac{ AB }{ BC }=\frac{12 k }{35 k }=\frac{12}{35}$
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
Simplify the following:
$175 \times 175+2 \times 175 \times 25+25 \times 25$
→$175 \times 175+2 \times 175 \times 25+25 \times 25$
Factorize:$\text{x}^2-2\sqrt{2}\text{x}-30$
→Find the mode for the following series:
7.5, 7.3, 7.2, 7.2, 7.4, 7.7, 7.7, 7.5, 7.3, 7.2, 7.6, 7.2
7.5, 7.3, 7.2, 7.2, 7.4, 7.7, 7.7, 7.5, 7.3, 7.2, 7.6, 7.2
Prove the Theorem: The measure of an exterior angle of a triangle is equal to the sum of its remote interior angles.
Find the curved surface area of a cone, if its slant height is 60cm and the radius of its base is 21cm.
Solve the following simultaneous equations $:2x + y = -2 ; 3x – y = 7$
→Give an example of a number $x$ such that $x^2$ is an irrational number and $x^3$ is a rational number.
→Measuring height of a tree using trigonometric ratios.
This experiment can be conducted on a clear sunny day. Look at the figure given above. Height of the tree is QR, height of the stick is BC.
Thrust a stick in the ground as shown in the figure. Measure its height and length of its shadow. Also measure the length of the shadow of the tree. Using these values, how will you determine the height of the tree?
Express the following equation in the form ax + by + c = 0 and indicate the values of a, b, c in case.
4x = 5y
4x = 5y
Solve the following equations: $4^{\text{x}-1}\times(0.5)^{3-2\text{x}}=\Big(\frac{1}{8}\Big)^{\text{x}}$
→