In triangle ABC, right angled at B if sin A = 12, find the value of (a). sin C cos A – cos C sin A (b) cos A cos C + sin A sin C
Answers
Answer:
Let ΔABC in which ∠B=90°
tan A = BC/AB = 1/√3
Let us consider
BC = 1k
AB = √3 k
Where k is the positive real number of the problem
By Pythagoras theorem in ΔABC we get:
AC2 = AB2+ BC2
AC2 = (√3 k)2 + (k)2
AC 2= 3k2 + k2
AC2 = 4k2
AC = 2k
Now find the values of cos A, Sin A
Sin A = BC/AC = 1/2
Cos A = AB/AC = √3/2
Then find the values of cos C and sin C
Sin C = AB/AC = √3/2
Cos C = BC/AC = 1/2
Now, substitute the values in the given problem
(i) sin A cos C + cos A sin C
⇒ (1/2) ×(1/2 )+ √3/2 ×√3/2
⇒ 1/4 + 3/4
sin A cos C + cos A sin C = 1
(ii) cos A cos C – sin A sin C
⇒ (√3/2 )(1/2) – (1/2) (√3/2 )
cos A cos C – sin A sin C = 0
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Answer:
In ∆ ABC, <B =90°, are given.
Then,
10cosA.sinC.secA = 6
=>cosA. sinC. 1/cosA =6/10
sinC = 3/5
As we know that the sum of all angles of A triangle is 180°.
So, <A+ <B + <C =180°
=> <A + <C + 90° = 180°,[ <B = 90°, are given. ]
=> <A + <C = 180°–90°
=> <A = 90°– <C
By taking sine on both side, we get
=>sinA = sin(90°–C)
=>sinA = cosC,[ because sin(90°–∅) =cos∅ ]
=>sinA =√(1–sin²C),[because cos∅=√(1–sin²∅)]
=>sinA= √(1– 3²/5²)
=>sinA =√(25–9/25)
=>sinA =√16/25
sinA =4/5, Ans.
I hope it helps you…!!!
Step-by-step explanation:
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