Question

In ΔABC, right angle is at B, AB = 5 cm and ∠ACB = 30°. Determine the lengths of the sides BC and AC.

Answer 1

It is given that ,

In ∆ABC , <B = 90°

<ACB = 30°

i ) tan <ACB = AB/BC

tan 30° = 5/BC

( 1/√3 ) = 5/BC

BC = 5√3

ii ) Sin <ACB = AB/AC

sin 30° = 5/AC

1/2 = 5/AC

AC = 5 × ( 2/1 )

AC = 10

Therefore ,

AC = 10 and BC = 5/√3

I hope this helps you.

: )

Answer 2

$\sf\circ \: {One \: angle \: is \: given \: and \: two}\\ \sf{sides \: are \: not \: given.}$

$\sf\circ \: {You \: cannot \: use \: Pythagoras \: theorem} \: \red✘$

$\sf\circ \: {Use \: trigonometric \: ratios} \: \red✔$

${\color{orange}{\textsf{\textbf {❥ {\underline{\underline{Solution:}}}}}}}$

$\frak{Sin \: 30=\frac{opp \: side}{hyp}}$

$\sf\frac{1}{2} = \frac{AB}{AC}\\ \\\sf\frac{1}{2} = \frac{5}{AC} \\\\{\underline{\boxed{\bf\red{AC = 10\:cm}}}}$

$\sf\implies(how \: to \: find \: BC \: ? \: you \: can \: either \: use \\ \sf Pythagoras \: theorem \: or \: trigonometric \: ratios)$

$\\\frak{Cos \: 30=\frac{adjacent \: side}{Hyp}}$

$\sf\frac{\sqrt{3} }{2} = \frac{BC}{AC}$

$\sf\frac{\sqrt{3}}{2} = \frac{BC}{10}$

$\sf{2 \times BC = 10\sqrt{3}}$

$\sf \: BC =\frac{{\cancel{10}^{ \: \red5} } \sqrt{3} }{\cancel2}$

${\underline{\boxed{\bf\red{BC = 5 \sqrt{3}}}}}$