Sainik School Previous Year Question Paper 2024 | Sainik School Maths |Sainik School Online Coaching

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Sainik School Coaching & RMS By - Sukhoi Academy

Sainik School Coaching & RMS By - Sukhoi Academy

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@akhandgamer1142
@akhandgamer1142 2 сағат бұрын
First comment Answer is 1 - 33°, 55°, 92° Explanation : 16x+4=180° x=176/16 x=11 Therefore, 3x=33° 5x=55° 8x+4=92° My name is Divyanshu Tiwari sir
@AvniSharma9271
@AvniSharma9271 2 сағат бұрын
challege question answer is aaaa 33,55,92 and you are a great teacher
@RamveerSingh-e9g
@RamveerSingh-e9g 2 сағат бұрын
Answer😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊Thanos ka jalva
@I_AM_VINEET_TIWARI
@I_AM_VINEET_TIWARI 2 сағат бұрын
Aaa, ❤❤❤❤❤
@savitabhatt8235
@savitabhatt8235 2 сағат бұрын
NAME - AYUSH BHATT ANSWER OF THE CHALLENGING QUESTION IS - OPTION A. 33° 55° and 92°
@Ovee_artcraft
@Ovee_artcraft 2 сағат бұрын
Name- Veer
@Ovee_artcraft
@Ovee_artcraft 2 сағат бұрын
Challenging question Ans - option a 33 , 55 , 92 x = 11
@Utkarsh708ww
@Utkarsh708ww 2 сағат бұрын
Ans. 33,55,92. Utkarsh singh
@Utkarsh708ww
@Utkarsh708ww 2 сағат бұрын
Ans 33,55,92. UTKARSH SINGH RAJPUT
@CK_KHUSH_BHATI
@CK_KHUSH_BHATI 2 сағат бұрын
CHALLENGE ANS- AAA FIRST SIR ME..!🥇🥇🥇🥇🥇🥇🥇🥇🥇🥇🥇🥇🥇🥇🥇🥇🥇🥇🥇🥇🥇🥇🥇🥇🥇🥇🥇🥇🥇🥇🥇🥇🥇🥇🥇🥇
@CK_KHUSH_BHATI
@CK_KHUSH_BHATI 2 сағат бұрын
ALL SEE SECS I AM FIRST
@AmNormalBoy
@AmNormalBoy 2 сағат бұрын
BRO WHEN I COMMENTED I DON'T SEE ANYONE IN COMMENTS 😅😊 ❤
@YuvaanSharma-tv9ip
@YuvaanSharma-tv9ip 2 сағат бұрын
answer to this question is option-a. Sir you are my hero
@subhashvandu6052
@subhashvandu6052 2 сағат бұрын
My name is - Suryansh Challenging question answer is --- option (A) 33,55,92
@Utkarsh708ww
@Utkarsh708ww 2 сағат бұрын
Name. Utkarsh singh rajput. Correct ans option a. 33,55,92
@AmNormalBoy
@AmNormalBoy 2 сағат бұрын
@dipendrasingh7818
@dipendrasingh7818 2 сағат бұрын
CHAALLENGING QUESTION = (A) 33, 55, 92
@Utkarsh708ww
@Utkarsh708ww 2 сағат бұрын
My ans is 33,55,92. Utkarsh singh
@Hemraj.education
@Hemraj.education 2 сағат бұрын
A wala button 👇👇👇👇👇👇👇👇👇👇👇
@abhinavpuri1283
@abhinavpuri1283 2 сағат бұрын
CHALLANGE Answer.AAAAAAAAAAAA
@rajeshdigrwal2228
@rajeshdigrwal2228 2 сағат бұрын
3X , 5X, 8X+4 = 180 16X +4 = 180 16X = 180 - 4 X =176/16 X=11 3X=33 5X=55 8X + 4 = 88+4 =92 ANSWER =A 33,55, 92
@ShivanshDwivedi-p3r
@ShivanshDwivedi-p3r Сағат бұрын
50/50
@arifaashraf152
@arifaashraf152 Сағат бұрын
Name: jannat zehra Class:6 Answer:aaaa 33,55,92
@SamriddhiRMS
@SamriddhiRMS 2 сағат бұрын
THE ANSWER OF THE CHALLENGING QUESTION IS OPTION NO. A ( 33° ,55° , 92° ) 👍 LOVE YOU SIR
@Saurabh1-d6t
@Saurabh1-d6t 2 сағат бұрын
my answer of challenge ahead is a 33° 55° 92° ❤ Solution We know that the sum of the angles in a triangle is 180 degrees. So, we can set up the following equation: 3x + 5x + (8x + 4) = 180 Combining like terms: 16x + 4 = 180 Subtracting 4 from both sides: 16x = 176 Dividing both sides by 16: x = 11 Now we can find the measure of each angle: * Angle A = 3x = 3 * 11 = 33 degrees * Angle B = 5x = 5 * 11 = 55 degrees * Angle C = 8x + 4 = 8 * 11 + 4 = 88 + 4 = 92 degrees Therefore, the measures of the angles are: * Angle A = 33 degrees * Angle B = 55 degrees * Angle C = 92 degrees
@CK_KHUSH_BHATI
@CK_KHUSH_BHATI 2 сағат бұрын
FROM CHATGPT COPIED THE ANSWER..!
@Saurabh1-d6t
@Saurabh1-d6t 2 сағат бұрын
No i have edited the reaply
@ShivanshDwivedi-p3r
@ShivanshDwivedi-p3r Сағат бұрын
D
@esportgaming8681
@esportgaming8681 2 сағат бұрын
3x+5x+8x+4=180 16x=176 X=11 Angles=33,55,92(A)
@Drpriyanka_3005
@Drpriyanka_3005 2 сағат бұрын
My name is Aaradhya Kulkarni. Challenge accepted
@abhijitpawar703
@abhijitpawar703 2 сағат бұрын
*NAME-SWARAJ* *ANSWER-A)33°,55°,92°* *THANK YOU SIR 🙂*
@JanviTyagi-y6e
@JanviTyagi-y6e 2 сағат бұрын
Sir correct answer will be 33° 55° 92°
@JanviTyagi-y6e
@JanviTyagi-y6e 2 сағат бұрын
Option A
@PrateekKumar-i1h
@PrateekKumar-i1h 2 сағат бұрын
Sir challenge questions answer is option a Angle a = 33 Angle b = 55 Angle c = 92
@Drpriyanka_3005
@Drpriyanka_3005 2 сағат бұрын
My name is Aaradhya Kulkarni. My answer is option A. B cuz x=11 So 3x=33 5x=55 8x+4=92
@sarjityadav6895
@sarjityadav6895 2 сағат бұрын
Challenge accepted Ans = A (33,55,92) Aryan Yadav Sol = 3x + 5x+ (8x + 4 ) = 180* 16 x + 4 = 180* 16x = 180 - 4 = 176 x = 11 3x = 33* 5x = 55* 8x + 4 = 92*
@chiragsharma1746
@chiragsharma1746 2 сағат бұрын
Chirag sharma I am from rajasthan Sir the chaellaemging question ans is angle,abc 33०,55०,92० Rms
@meenuchaudhary8190
@meenuchaudhary8190 2 сағат бұрын
a option is right answer my name is arjun chaudhary
@ShivanshDwivedi-p3r
@ShivanshDwivedi-p3r Сағат бұрын
C
@AbhinavMishra-ii2hg
@AbhinavMishra-ii2hg 2 сағат бұрын
Sir my answer a option and i love you sir
@ttrkmotivation
@ttrkmotivation 2 сағат бұрын
A is the ans sir I am tannu
@riturana5548
@riturana5548 Сағат бұрын
Good evening sir Sir I am Priyanshi . Answer to its question is option a 33° , 55° , 92°
@ReenaDevi-t6t
@ReenaDevi-t6t 2 сағат бұрын
ANSWER IS A
@dipendrasingh7818
@dipendrasingh7818 2 сағат бұрын
ANSWER IS OPTION A I TAKE THE CHALLENGE MY NAME IS ARYAN SINGH
@sikandar1974
@sikandar1974 2 сағат бұрын
Sar my answer is 33°55°92°
@rajeshdigrwal2228
@rajeshdigrwal2228 2 сағат бұрын
CHALLANGE ANSWER = A I AM LIVE IN HARYANA
@arabhmehraclass3b714
@arabhmehraclass3b714 2 сағат бұрын
Aaaaaaaaaaaaa I am Arabh
@RbindraSingh46
@RbindraSingh46 2 сағат бұрын
Aaaa (33,55,92)🎉🎉🎉🎉🎉🎉 sir please shoutout my name please ✊✊✊
@YuvaanSharma-tv9ip
@YuvaanSharma-tv9ip 2 сағат бұрын
SIR ANSWER TO THE CHALLENGING QUESTION IS OPTION-A THAT IS 33,55,92. EXPLANATION- 3X+5X+8X+4=180 , 16x+4=180 16x= 176, x= 176/16= 11, then to find out angles we will put the value of x and multiply it.
@AadityaRaj-l1s
@AadityaRaj-l1s 2 сағат бұрын
My name is Aaditya Raj . I am from Palwal near Faridabad . My Aim is To crack NDA and fly Sukhoi 30 MKI fighter jet . MY DREAM SCHOOL IS RMS CHAIL / SAINIK SCHOOL KUNNJPURA . ANSWER OF THE CHALLENGING QUESTION IS OPTION NUMBER (A) 🇮🇳JAI HIND JAI BHARAT 🇮🇳🫡🫡🫡🫡
@Ovee_artcraft
@Ovee_artcraft 2 сағат бұрын
Sir don't ignore me
@Socialkhoj1
@Socialkhoj1 2 сағат бұрын
Sir my name is Shubham Yadav ,,, I have accepted your challenging question and my answer is {{a}} 33, 55, 92..... thanks
@Deveshkumarsharmabhel
@Deveshkumarsharmabhel 2 сағат бұрын
My name is Aayu Sharma from UP. ANS is (33 degree, 55 degree and 92 degree) nam lelena sir
@Dr.smasher6
@Dr.smasher6 2 сағат бұрын
Good Evening sir My name is Manvik Mishra Answers. Challenging question -(A) Sir you are the best teacher in the world ❤❤❤❤❤ Please take my name 😊😊😊😊
@nishithkumar-l6u
@nishithkumar-l6u 2 сағат бұрын
name = NISHITH KUMAR answer is (1) or (a) 33,55,92 To find the angles \(A\), \(B\), and \(C\) in triangle \(ABC\), given that: (A = 3X\) (B = 5X\) (C = 8X + 4\) we can use the fact that the sum of the angles in a triangle is \(180^\circ\): \[ A + B + C = 180^\circ \] Substituting the expressions for \(A\), \(B\), and \(C\) into this equation gives: \[ 3X + 5X + (8X + 4) = 180 \] Now, combine the like terms: \[ (3X + 5X + 8X) + 4 = 180 \] \[ 16X + 4 = 180 \] Next, subtract \(4\) from both sides: \[ 16X = 180 - 4 \] \[ 16X = 176 \] Now, divide by \(16\) to solve for \(X\): \[ X = \frac {176}{16} = 11 \] Now that we have \(X\), we can find each angle: 1. **Angle A: ** \[ A = 3X = 3 \cdot 11 = 33^\circ \] 2. **Angle B: ** \[ B = 5X = 5 \cdot 11 = 55^\circ \] 3. **Angle C:** \[ C = 8X + 4 = 8 \cdot 11 + 4 = 88 + 4 = 92^\circ \] Thus, the angles of triangle \(ABC\) are: (A = 33^\circ\) (B = 55^\circ\) (C = 92^\circ\) To verify: 33 + 55 + 92 = 180^\circ
@arabhmehraclass3b714
@arabhmehraclass3b714 2 сағат бұрын
Aaaaaaaaaaaaa I am Arabh
@nishithkumar-l6u
@nishithkumar-l6u 2 сағат бұрын
name = NISHITH KUMAR answer is (1) or (a) 33,55,92 To find the angles \(A\), \(B\), and \(C\) in triangle \(ABC\), given that: (A = 3X\) (B = 5X\) (C = 8X + 4\) we can use the fact that the sum of the angles in a triangle is \(180^\circ\): \[ A + B + C = 180^\circ \] Substituting the expressions for \(A\), \(B\), and \(C\) into this equation gives: \[ 3X + 5X + (8X + 4) = 180 \] Now, combine the like terms: \[ (3X + 5X + 8X) + 4 = 180 \] \[ 16X + 4 = 180 \] Next, subtract \(4\) from both sides: \[ 16X = 180 - 4 \] \[ 16X = 176 \] Now, divide by \(16\) to solve for \(X\): \[ X = \frac {176}{16} = 11 \] Now that we have \(X\), we can find each angle: 1. **Angle A: ** \[ A = 3X = 3 \cdot 11 = 33^\circ \] 2. **Angle B: ** \[ B = 5X = 5 \cdot 11 = 55^\circ \] 3. **Angle C:** \[ C = 8X + 4 = 8 \cdot 11 + 4 = 88 + 4 = 92^\circ \] Thus, the angles of triangle \(ABC\) are: (A = 33^\circ\) (B = 55^\circ\) (C = 92^\circ\) To verify: 33 + 55 + 92 = 180^\circ
@nishithkumar-l6u
@nishithkumar-l6u 2 сағат бұрын
name = NISHITH KUMAR answer is (1) or (a) 33,55,92 To find the angles \(A\), \(B\), and \(C\) in triangle \(ABC\), given that: (A = 3X\) (B = 5X\) (C = 8X + 4\) we can use the fact that the sum of the angles in a triangle is \(180^\circ\): \[ A + B + C = 180^\circ \] Substituting the expressions for \(A\), \(B\), and \(C\) into this equation gives: \[ 3X + 5X + (8X + 4) = 180 \] Now, combine the like terms: \[ (3X + 5X + 8X) + 4 = 180 \] \[ 16X + 4 = 180 \] Next, subtract \(4\) from both sides: \[ 16X = 180 - 4 \] \[ 16X = 176 \] Now, divide by \(16\) to solve for \(X\): \[ X = \frac {176}{16} = 11 \] Now that we have \(X\), we can find each angle: 1. **Angle A: ** \[ A = 3X = 3 \cdot 11 = 33^\circ \] 2. **Angle B: ** \[ B = 5X = 5 \cdot 11 = 55^\circ \] 3. **Angle C:** \[ C = 8X + 4 = 8 \cdot 11 + 4 = 88 + 4 = 92^\circ \] Thus, the angles of triangle \(ABC\) are: (A = 33^\circ\) (B = 55^\circ\) (C = 92^\circ\) To verify: 33 + 55 + 92 = 180^\circ
@nishithkumar-l6u
@nishithkumar-l6u 2 сағат бұрын
name = NISHITH KUMAR answer is (1) or (a) 33,55,92 To find the angles \(A\), \(B\), and \(C\) in triangle \(ABC\), given that: (A = 3X\) (B = 5X\) (C = 8X + 4\) we can use the fact that the sum of the angles in a triangle is \(180^\circ\): \[ A + B + C = 180^\circ \] Substituting the expressions for \(A\), \(B\), and \(C\) into this equation gives: \[ 3X + 5X + (8X + 4) = 180 \] Now, combine the like terms: \[ (3X + 5X + 8X) + 4 = 180 \] \[ 16X + 4 = 180 \] Next, subtract \(4\) from both sides: \[ 16X = 180 - 4 \] \[ 16X = 176 \] Now, divide by \(16\) to solve for \(X\): \[ X = \frac {176}{16} = 11 \] Now that we have \(X\), we can find each angle: 1. **Angle A: ** \[ A = 3X = 3 \cdot 11 = 33^\circ \] 2. **Angle B: ** \[ B = 5X = 5 \cdot 11 = 55^\circ \] 3. **Angle C:** \[ C = 8X + 4 = 8 \cdot 11 + 4 = 88 + 4 = 92^\circ \] Thus, the angles of triangle \(ABC\) are: (A = 33^\circ\) (B = 55^\circ\) (C = 92^\circ\) To verify: 33 + 55 + 92 = 180^\circ
@nishithkumar-l6u
@nishithkumar-l6u 2 сағат бұрын
name = NISHITH KUMAR answer is (1) or (a) 33,55,92 To find the angles \(A\), \(B\), and \(C\) in triangle \(ABC\), given that: (A = 3X\) (B = 5X\) (C = 8X + 4\) we can use the fact that the sum of the angles in a triangle is \(180^\circ\): \[ A + B + C = 180^\circ \] Substituting the expressions for \(A\), \(B\), and \(C\) into this equation gives: \[ 3X + 5X + (8X + 4) = 180 \] Now, combine the like terms: \[ (3X + 5X + 8X) + 4 = 180 \] \[ 16X + 4 = 180 \] Next, subtract \(4\) from both sides: \[ 16X = 180 - 4 \] \[ 16X = 176 \] Now, divide by \(16\) to solve for \(X\): \[ X = \frac {176}{16} = 11 \] Now that we have \(X\), we can find each angle: 1. **Angle A: ** \[ A = 3X = 3 \cdot 11 = 33^\circ \] 2. **Angle B: ** \[ B = 5X = 5 \cdot 11 = 55^\circ \] 3. **Angle C:** \[ C = 8X + 4 = 8 \cdot 11 + 4 = 88 + 4 = 92^\circ \] Thus, the angles of triangle \(ABC\) are: (A = 33^\circ\) (B = 55^\circ\) (C = 92^\circ\) To verify: 33 + 55 + 92 = 180^\circ
@nishithkumar-l6u
@nishithkumar-l6u 2 сағат бұрын
name = NISHITH KUMAR answer is (1) or (a) 33,55,92 To find the angles \(A\), \(B\), and \(C\) in triangle \(ABC\), given that: (A = 3X\) (B = 5X\) (C = 8X + 4\) we can use the fact that the sum of the angles in a triangle is \(180^\circ\): \[ A + B + C = 180^\circ \] Substituting the expressions for \(A\), \(B\), and \(C\) into this equation gives: \[ 3X + 5X + (8X + 4) = 180 \] Now, combine the like terms: \[ (3X + 5X + 8X) + 4 = 180 \] \[ 16X + 4 = 180 \] Next, subtract \(4\) from both sides: \[ 16X = 180 - 4 \] \[ 16X = 176 \] Now, divide by \(16\) to solve for \(X\): \[ X = \frac {176}{16} = 11 \] Now that we have \(X\), we can find each angle: 1. **Angle A: ** \[ A = 3X = 3 \cdot 11 = 33^\circ \] 2. **Angle B: ** \[ B = 5X = 5 \cdot 11 = 55^\circ \] 3. **Angle C:** \[ C = 8X + 4 = 8 \cdot 11 + 4 = 88 + 4 = 92^\circ \] Thus, the angles of triangle \(ABC\) are: (A = 33^\circ\) (B = 55^\circ\) (C = 92^\circ\) To verify: 33 + 55 + 92 = 180^\circ
@nishithkumar-l6u
@nishithkumar-l6u 2 сағат бұрын
name = NISHITH KUMAR answer is (1) or (a) 33,55,92 To find the angles \(A\), \(B\), and \(C\) in triangle \(ABC\), given that: (A = 3X\) (B = 5X\) (C = 8X + 4\) we can use the fact that the sum of the angles in a triangle is \(180^\circ\): \[ A + B + C = 180^\circ \] Substituting the expressions for \(A\), \(B\), and \(C\) into this equation gives: \[ 3X + 5X + (8X + 4) = 180 \] Now, combine the like terms: \[ (3X + 5X + 8X) + 4 = 180 \] \[ 16X + 4 = 180 \] Next, subtract \(4\) from both sides: \[ 16X = 180 - 4 \] \[ 16X = 176 \] Now, divide by \(16\) to solve for \(X\): \[ X = \frac {176}{16} = 11 \] Now that we have \(X\), we can find each angle: 1. **Angle A: ** \[ A = 3X = 3 \cdot 11 = 33^\circ \] 2. **Angle B: ** \[ B = 5X = 5 \cdot 11 = 55^\circ \] 3. **Angle C:** \[ C = 8X + 4 = 8 \cdot 11 + 4 = 88 + 4 = 92^\circ \] Thus, the angles of triangle \(ABC\) are: (A = 33^\circ\) (B = 55^\circ\) (C = 92^\circ\) To verify: 33 + 55 + 92 = 180^\circ
@nishithkumar-l6u
@nishithkumar-l6u 2 сағат бұрын
name = NISHITH KUMAR answer is (1) or (a) 33,55,92 To find the angles \(A\), \(B\), and \(C\) in triangle \(ABC\), given that: (A = 3X\) (B = 5X\) (C = 8X + 4\) we can use the fact that the sum of the angles in a triangle is \(180^\circ\): \[ A + B + C = 180^\circ \] Substituting the expressions for \(A\), \(B\), and \(C\) into this equation gives: \[ 3X + 5X + (8X + 4) = 180 \] Now, combine the like terms: \[ (3X + 5X + 8X) + 4 = 180 \] \[ 16X + 4 = 180 \] Next, subtract \(4\) from both sides: \[ 16X = 180 - 4 \] \[ 16X = 176 \] Now, divide by \(16\) to solve for \(X\): \[ X = \frac {176}{16} = 11 \] Now that we have \(X\), we can find each angle: 1. **Angle A: ** \[ A = 3X = 3 \cdot 11 = 33^\circ \] 2. **Angle B: ** \[ B = 5X = 5 \cdot 11 = 55^\circ \] 3. **Angle C:** \[ C = 8X + 4 = 8 \cdot 11 + 4 = 88 + 4 = 92^\circ \] Thus, the angles of triangle \(ABC\) are: (A = 33^\circ\) (B = 55^\circ\) (C = 92^\circ\) To verify: 33 + 55 + 92 = 180^\circ
@nishithkumar-l6u
@nishithkumar-l6u 2 сағат бұрын
name = NISHITH KUMAR answer is (1) or (a) 33,55,92 To find the angles \(A\), \(B\), and \(C\) in triangle \(ABC\), given that: (A = 3X\) (B = 5X\) (C = 8X + 4\) we can use the fact that the sum of the angles in a triangle is \(180^\circ\): \[ A + B + C = 180^\circ \] Substituting the expressions for \(A\), \(B\), and \(C\) into this equation gives: \[ 3X + 5X + (8X + 4) = 180 \] Now, combine the like terms: \[ (3X + 5X + 8X) + 4 = 180 \] \[ 16X + 4 = 180 \] Next, subtract \(4\) from both sides: \[ 16X = 180 - 4 \] \[ 16X = 176 \] Now, divide by \(16\) to solve for \(X\): \[ X = \frac {176}{16} = 11 \] Now that we have \(X\), we can find each angle: 1. **Angle A: ** \[ A = 3X = 3 \cdot 11 = 33^\circ \] 2. **Angle B: ** \[ B = 5X = 5 \cdot 11 = 55^\circ \] 3. **Angle C:** \[ C = 8X + 4 = 8 \cdot 11 + 4 = 88 + 4 = 92^\circ \] Thus, the angles of triangle \(ABC\) are: (A = 33^\circ\) (B = 55^\circ\) (C = 92^\circ\) To verify: 33 + 55 + 92 = 180^\circ
@nishithkumar-l6u
@nishithkumar-l6u 2 сағат бұрын
name = NISHITH KUMAR answer is (1) or (a) 33,55,92 To find the angles \(A\), \(B\), and \(C\) in triangle \(ABC\), given that: (A = 3X\) (B = 5X\) (C = 8X + 4\) we can use the fact that the sum of the angles in a triangle is \(180^\circ\): \[ A + B + C = 180^\circ \] Substituting the expressions for \(A\), \(B\), and \(C\) into this equation gives: \[ 3X + 5X + (8X + 4) = 180 \] Now, combine the like terms: \[ (3X + 5X + 8X) + 4 = 180 \] \[ 16X + 4 = 180 \] Next, subtract \(4\) from both sides: \[ 16X = 180 - 4 \] \[ 16X = 176 \] Now, divide by \(16\) to solve for \(X\): \[ X = \frac {176}{16} = 11 \] Now that we have \(X\), we can find each angle: 1. **Angle A: ** \[ A = 3X = 3 \cdot 11 = 33^\circ \] 2. **Angle B: ** \[ B = 5X = 5 \cdot 11 = 55^\circ \] 3. **Angle C:** \[ C = 8X + 4 = 8 \cdot 11 + 4 = 88 + 4 = 92^\circ \] Thus, the angles of triangle \(ABC\) are: (A = 33^\circ\) (B = 55^\circ\) (C = 92^\circ\) To verify: 33 + 55 + 92 = 180^\circ
@nishithkumar-l6u
@nishithkumar-l6u 2 сағат бұрын
name = NISHITH KUMAR answer is (1) or (a) 33,55,92 To find the angles \(A\), \(B\), and \(C\) in triangle \(ABC\), given that: (A = 3X\) (B = 5X\) (C = 8X + 4\) we can use the fact that the sum of the angles in a triangle is \(180^\circ\): \[ A + B + C = 180^\circ \] Substituting the expressions for \(A\), \(B\), and \(C\) into this equation gives: \[ 3X + 5X + (8X + 4) = 180 \] Now, combine the like terms: \[ (3X + 5X + 8X) + 4 = 180 \] \[ 16X + 4 = 180 \] Next, subtract \(4\) from both sides: \[ 16X = 180 - 4 \] \[ 16X = 176 \] Now, divide by \(16\) to solve for \(X\): \[ X = \frac {176}{16} = 11 \] Now that we have \(X\), we can find each angle: 1. **Angle A: ** \[ A = 3X = 3 \cdot 11 = 33^\circ \] 2. **Angle B: ** \[ B = 5X = 5 \cdot 11 = 55^\circ \] 3. **Angle C:** \[ C = 8X + 4 = 8 \cdot 11 + 4 = 88 + 4 = 92^\circ \] Thus, the angles of triangle \(ABC\) are: (A = 33^\circ\) (B = 55^\circ\) (C = 92^\circ\) To verify: 33 + 55 + 92 = 180^\circ
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