Q. 8 A4.2( 27 Votes )

# In a right angle

We have

Given that, tan A = √3/1

And tan A is given by,

perpendicular = √3x and base = x

Then, we can use Pythagoras theorem in ∆ABC,

(hypotenuse)2 = (perpendicular)2 + (base)2

(hypotenuse)2 = (√3x)2 + (x)2

(hypotenuse)2 = 3x2 + x2 = 4x2

hypotenuse = √(4x2) = 2x

We have, AB = √3x, BC = x and AC = 2x.

Using these values,

…(i)

Similarly,

…(ii)

Also,

…(iii)

Similarly,

…(iv)

We have to solve: sin A cos C + cos A sin C.

Substituting equations (i), (ii), (iii) & (iv) in above,

sin A cos C + cos A sin C =

sin A cos C + cos A sin C = 1/4 + 3/4

sin A cos C + cos A sin C = 4/4 = 1

Thus, sin A cos C + cos A sin C = 1.

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