Question

in ∆LMN right angled at M LM =3cm,MN=cm and LN =5cm determine all six trigonometric ratio of angle L​

Answer 1

$\star\:\:\:\bf\large\underline\blue{Question:-}$

• Find all the six T-Ratio for angle L.

$\star\:\:\:\bf\large\underline\blue{Solution:-}$

Before we start, please check out the attachment.

Here,

$LM=3cm$

$LN=5cm$

By pythagoras theorem,

$LN^{2}=LM^{2}+MN^{2}$

So,

$MN=\sqrt{LN^{2}-LM^{2}}$

$= \sqrt{ {5}^{2} - {3}^{2} }$

$= \sqrt{25 - 9}$

$= \sqrt{16}$

$= 4cm$

So,

$\sin \alpha = \frac{4}{5}$

$\cos \alpha = \frac{3}{5}$

$\tan \alpha = \frac{4}{3}$

$\cot \alpha = \frac{3}{4}$

$\sec \alpha = \frac{5}{3}$

$\cosec \alpha = \frac{5}{4}$

Answer 2

Answer:

Here are steps of construction :

First of all draw a line segment MN = 3 cm

Using a compass draw an angle of 90° at M i.e. <NMQ =90°

(Using Pythagoras theorem we get LM = 4 cm )

Mark an arc on Ray MQ taking the radius of compass = 4 cm, name the point of intersection as L

Join L to N

∆LMN is the required triangle