Question

pls answer fast...... class XI trigonometry.......

Answer 1

We know that In a Triangle the angles give up sum of 180°, Then

A + B + C = 180°         

A = 180° - B - C           .............(1)

Also from question,

cos(B+C-A) = 0       

From the trigonometric table cos 90° = 0

cos(B+C-A) = cos 90°

Cancelling cos on both sides 

B + C - A = 90°             ...............(2)

Similarly 

sin(C+A-B) = √3/2

From the trigonometric table sin 60° = √3/2

sin(C+A-B) = sin 60°

Cancelling sin on both sides

C + A - B = 60°             .................(3)
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From (1) on substituting it in (2) and (3)

(2) ⇒ B + C - (180°-B-C) = 90°

          B + C - 180° + B + C = 90°

          2 B + 2 C = 270°

                B + C = 135°      ..................(4)

Also,

(3) ⇒  C + 180° - B - C - B = 60°

                                    2 B = 180° - 60°
 
                                       B = 120°/2

                                       B = 60°    ..................(5)

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Substituting (5) in (4) and (1)

(4) ⇒ B + C = 135°

                C = 135° - 60°

                C = 75°

(1) ⇒ A = 180° - B - C

          A = 180° - 60° - 75°

          A = 45°
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Hence the angles are A = 45° B =120° C = 75°
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