If ∆ABC is right angled at C, then the value of cos (A+B) is
a. 0
b. 1
c. 1/2
d. √3/2

Solution:

Given, the triangle ABC is a right triangle with C at right angle.

We have to find the value of cos(A+B)

If ∆ABC is right angled at C, then the value of cos (A+B) is

We know that the sum of all the three interior angles of a triangle is always equal to 180°

So, ∠A + ∠B + ∠C = 180°

∠A + ∠B + 90° = 180°

∠A + ∠B = 180° - 90°

∠A + ∠B = 90°

So, cos (A+B) = cos 90°

Using the trigonometric ratios of angles,

cos 90° = 0

Therefore, the value of cos(A + B) is zero.

✦ Try This: If ∆DEF is right angled at E, then the value of sin (D + F) is

Given, the triangle DEF is a right triangle with E at right angles.

We have to find the value of sin(D + F)

We know that the sum of all the three interior angles of a triangle is always equal to 180°

So, ∠D + ∠E + ∠F = 180°

∠D + ∠F + 90° = 180°

∠D + ∠F = 180° - 90°

∠D + ∠F = 90°

So, sin (D + F) = sin 90°

Using the trigonometric ratio of angels,

sin 90° = 1

Therefore, the value of sin(D + F) is 1.

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 8

NCERT Exemplar Class 10 Maths Exercise 8.1 Problem 8

If ∆ABC is right angled at C, then the value of cos (A+B) is a. 0, b. 1, c. 1/2, d. √3/2

Summary:

If ∆ABC is right angled at C, then the value of cos (A+B) is 0

☛ Related Questions: