Question

CLASS 10
TRIGONOMETRY (HOTS)

NO SPAMMING
CONTENT QUALITY

Answer 1

In the give figure,
AD=4cm
BD=3cm
CB=12cm
Angle C=theta
AB^2=AD^2+BD^2
=4^2+3^2
=16+9
=25
AB=
$\sqrt{25}$
AB=5cm
So, cot theta=AB/CB
=5/12

Answer 2

$\mathfrak{Answer\: 1}$

Given, AD= 4 cm

BD= 3cm

CB= 12cm

$\angle$ ADB=90°

angle ABC= 90°

To find, cot$\theta$

So, first let's find the value of AB.

We know that $\angle$ ADB is a right angle.

So, AB can be found by Pythagoras theorem.

By Pythagoras theorem

$AB^2=AD^2+BD^2\\ AB^2=(4)^2+(3)^2\\ AB^2=16+9\\ AB^2=25\\ AB=\sqrt25\\ \therefore AB=5 cm$

Now, AC can be also find similarly by Pythagoras theorem.

Because, $\angle$ ABC= 90°

, $AC^2=AB^2+BC^2\\ AC^2=(5)^2+(12)^2\\ AC^2=25+144\\ AC^2=169\\ \therefore AC=13 cm$.

We know,

$cot\theta=\frac{adjescent}{opposite}\\ \\ =\frac{12}{5}$

$\mathfrak{Answer\: 2}$

$Given,\:tan\theta=\frac{4}{3}\\ \\ But\ we\ know, tan\theta=\frac{opposite}{adjescent}$

If $\theta$is the angle, then,

opposite=AB=5 cm

adjescent=BC=12 cm

hypotenuse=AC=13 cm

But we know, $sin\theta=\frac{opposite}{hypotenuse}\\ \\ cos\theta=\frac{adjescent}{hypotenuse}$

To find,

$\frac{sin\theta+cos\theta}{sin\theta-cos\theta}\\ \\Putting values of sin\theta and cos\theta$