Question

In triangle ABC,right angle at B,AB =24 cm,BC=7 cm.detrrmine, (i) sin A ,cos A(ii) sin C , cos C.

Answer 1

In ∆ ABC
B =90°

AB = 24 cm
BC = 7cm
Therefore hypotenuse AC will be
√24²+7²
= √576+49
=√ 625
=25cm

SinA = BC/AC
= 7/25
Cos A= AB/AC
= 24/25

sinC = AB/AC
= 24/25
CosC = BC/AC
= 7/25

Answer 2

In Δ ABC, right-angled at B

Using Pythagoras theorem

AC² = AB² +BC²  

AC² = 576 + 49 = 625

AC = √635

AC = ±25

$AC\:=\:+ 25$

Now

(i) In a right angle triangle ABC where B=90° ,

Sin A = $\frac{BC}{AC}$

= $\frac{7}{25}$

CosA = $\frac{AB}{AC}$

= $\frac{24}{25}$

(ii) Sin C = $\frac{AB}{AC}$

= $\frac{24}{25}$

Cos C = $\frac{BC}{AC}$

= $\frac{7}{25}$