Question

in the given figure angle ABC = _​

Answer 1

In the above question , the following information is given -

There is a right angled triangle ABC where -

Angle ACB = 90°

AC = 100 √ 3

BC = 100

To find -

Find the value of Angle ABC ..

Solution -

• Method 1 { Trigonometry ]

Here ,

We know that -

Tan theta = [ Perpendicular / Base ]

In angle ABC ,

Tan ABC = [ AC / CB ]

=> [ 100 √ 3 / 100 ]

=> √ 3

So ,

Tan ABC = √ 3

Angle ABC = tan inverse of √ 3 .

tan inverse of √ 3 is 60 °

So ,

Angle ABC = 60°

Thus , the required value of Angle ABC is 60°

____________

Additional Information -

Sin Theta = [ Perpendicular / Hypotenuse ]

Cos theta = [ Base / Hypotenuse ]

______

Answer 2

$\huge\mathbb\pink{Answer}$

»Given:

It is a right angled triangle.

So, angle ACB = 90°.

AC = 100√3 cm (perpendicular)

BC = 100 cm(base)

» To find:

Angle ABC

» Solution:

We know that,

${\purple{\boxed{tan \theta = \frac{perpendicular}{base}}}}$

by putting the values -

$\frac{100 \sqrt{3}}{100} = \sqrt{3}$

We know that,

Tan 60° = √3

Therefore,

the required angle ABC = 60°.

Extra information-

»Trigonometric values -

$\large\begin{tabular}{ | c | c | c |}\cline{1-3} Angle & cos \theta & sin\theta\\ \cline{1 - 3} 0 ^{\circ} & 1 & 0 \\\cline{1-3} 30 ^{\circ} & \frac{\sqrt{3}}{2} & \frac{1}{2} \\\cline{1-3} 45^{\circ} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\\cline{1-3} 60 ^{\circ} & \frac{1}{2} & \frac{\sqrt{3}}{2} \\\cline{1-3}90 ^{\circ}& 0 & 1\\\cline{1-3}\end{tabular}$

* Trigonometric ratios -

$sin\theta = \frac{P}{H}\\ \\ \\ cos\theta= \frac{B}{H}\\ \\ \\ tan\theta = \frac{P}{B} / \frac{sin\theta}{cos\theta}\\ \\ \\cosec\theta = \frac{H}{P} / \frac{1}{sin\theta}\\ \\ \\ sec\theta = \frac{H}{B}/ \frac{1}{cos\theta}\\ \\ \\ cot\theta = \frac{B}{P}/\frac{1}{tan\theta}$

* B = Base

H = Hypotenuse

P = Perpendicular