Question

In a ∆ABC, right angled at A, if $tan C=\sqrt{3}$, find the value of sin B cos C + cos B sin C.

Answer 1

SOLUTION :  

Given:

In ΔABC , Right angled at A.  

tan C = √3

tan C = √3/1  = P/B = AB/AC

In right ∆ ABC, by using Pythagoras theorem

BC² =  AB² +AC²

BC² = (√3)² +  1²

BC²  = 3 + 1

BC² = 4

[by taking square root on both sides]

BC = √4

BC= 2

Hypotenuse side (BC)= 2  

With reference to ∠C , base = AC, Hypotenuse = BC , perpendicular = AB

With reference to ∠B , base = AB, Hypotenuse = BC,  perpendicular = AC

 Now, sin B = Perpendicular/ Hypotenuse

sin B= AC/ BC

sin B = 1/2  

Now, cos B= base / Hypotenuse

cos B = AB/BC

cos B= √3/2

sin C = Perpendicular/ Hypotenuse

sin C = AB/AC

sin C = √3/2

cos C= base / Hypotenuse

cos C  = AC/BC

cos C= 1/2

sinB cosC + cosB sin C

= ½ × 1/2 + √3/2 × √3/2

= ¼  + ¾  

= (1+3)/4

= 4/4

= 1

sinB cosC + cosB sin C  = 1

Hence, the value of sinB cosC + cosB sin C = 1

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