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christen
@christen
September 2021
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pls answer fast...... class XI trigonometry.......
Answers & Comments
nitthesh7
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)
________________________________________________________
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)
_________________________________________________________
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°
__________________________________________________________
Hence the angles are A = 45° B =120° C = 75°
__________________________________________________________
☺☺☺ Hope this Helps ☺☺☺
1 votes
Thanks 1
nitthesh7
sis r u 11th
christen
ya
christen
u?
nitthesh7
myself 10th going to step on class 11th
christen
ooo
christen
gd
nitthesh7
Hm... bye
christen
i hv just stepped in 11th class
christen
bye
nitthesh7
Tq for Brainliest sis
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Answers & Comments
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)
________________________________________________________
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)
_________________________________________________________
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°
__________________________________________________________
Hence the angles are A = 45° B =120° C = 75°
__________________________________________________________
☺☺☺ Hope this Helps ☺☺☺