Question

can someone tell me the solution of (b) in the image

Answer 1

Since TanA=1/✓3

So A =30

And by Pythagoras thm C=60

(i)1/2*1/2+✓3/2*✓3/2

=1/4+3/4

=4/4

=1

(ii)✓3/2*1/2-1/2*✓3/2

=✓3/4-✓3/4

=0

Answer 2

Trigonometric identities used,

$\bf \: Tan \: \theta \: = \: \frac{Perpendicular}{Base} \\ \\ \bf \: Sin \: \theta \: = \: \frac{Perpendicular}{Hypotenuse} \\ \\ \bf \: Cos \: \theta \: = \: \frac{Base}{Hypotenuse}$

Given,

A triangle ∆ABC is right angled at B

$Tan \: A \: = \frac{1}{ \sqrt{3} }$

We know that Tan theta = Perpendicular / Base

So, Perpendicular = 1
Base = √3

We can calculate the value of Hypotenuse by Pythagoras theorem.

${H}^{2} = {B}^{2} + {P}^{2} \\ \\ {H}^{2} = {(1)}^{2} + {( \sqrt{3} )}^{2} \\ \\ {H}^{2} = 1 + 3 \\ \\ H = \sqrt{4} \\ \\ H = 2cm$
$\bf \: (i) \: Sin \: A \: Cos \: C \: + \: Cos \: A \: Sin \: C \\ \\ = \frac{1}{2} \times \frac{1}{2} + \frac{ \sqrt{3} }{2} \times \frac{ \sqrt{3} }{2} \\ \\ = \frac{1}{4} + \frac{3}{4} \\ \\ = \frac{1 + 3}{4} \\ \\ = \frac{4}{4} \\ \\ = 1$
$\bf \: (ii) \: Cos \: A \: Cos \: C \: - \: Sin\: A \: Sin \: C \\ \\ = \frac{ \sqrt{3} }{2} \times \frac{1}{2} - \frac{1}{2} \times \frac{ \sqrt{3} }{2} \\ \\ = \frac{ \sqrt{3} }{4} - \frac{ \sqrt{3} }{4} \\ \\ = 0$

If any doubt, please ask ;)