How to solve the introduction to trigonometry question of ncert class 10 . Toady i am solving the exercise 8.1 question no. 9(1) its so hard for me to solve , please help me to solve this tricky question . In triangle ABC, right-angled at B, if tan A = 1/√3 find the value of: (i) sin A cos C + cos A sin C.
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In triangle ABC, right-angled at B, if tan A = 1/√3 find the value of: (i) sin A cos C + cos A sin C. Q.9(1)
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Let ΔABC in which ∠B=90°
tan A = BC/AB = 1/√3
Let BC = 1k and 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
AC2=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 = 1
(ii) cos A cos C – sin A sin C = (√3/2 )(1/2) – (1/2) (√3/2 ) = 0