MATH SOLVE

3 months ago

Q:
# in this right triangle LMN L and M are complentary angles and sin (L) is 19\20 what is cos (M)

Accepted Solution

A:

In any right triangle ABC, with right angle A we have the following:

[tex]\sin(\angle B)=\cos(\angle C)[/tex]

and

[tex]\sin(\angle C)=\cos(\angle B)[/tex].

This means that the sine of a non 90° angle, is equal to the cosine of its complementary angle.

Thus, [tex]\displaystyle{ \cos(\angle M)=\sin (\angle L)= \frac{19}{20} [/tex].

(Take a look at the picture. From the definition of trigonometric ratios in right triangles we have :

[tex]\displaystyle { \sin(\angle L)= \frac{opposite \ side}{hypotenuse} = \frac{19}{20}. [/tex]

and

[tex]\displaystyle { \cos(\angle M)= \frac{adjacent \ side}{hypotenuse} = \frac{19}{20}. [/tex])

Answer: 19/20

[tex]\sin(\angle B)=\cos(\angle C)[/tex]

and

[tex]\sin(\angle C)=\cos(\angle B)[/tex].

This means that the sine of a non 90° angle, is equal to the cosine of its complementary angle.

Thus, [tex]\displaystyle{ \cos(\angle M)=\sin (\angle L)= \frac{19}{20} [/tex].

(Take a look at the picture. From the definition of trigonometric ratios in right triangles we have :

[tex]\displaystyle { \sin(\angle L)= \frac{opposite \ side}{hypotenuse} = \frac{19}{20}. [/tex]

and

[tex]\displaystyle { \cos(\angle M)= \frac{adjacent \ side}{hypotenuse} = \frac{19}{20}. [/tex])

Answer: 19/20