Question

In the triangle ABC ,<B =900 AC=5cm , sin C =4/5
a) What is the length of AB?
b) Find cos C.​

Answer 1

${\huge{\rm{\underline{\underline{Question:-}}}}}$

Q) In a triangle ABC , $\angle$ B = 90° , AC = 5 cm , sin C = $\dfrac{4}{5}$

a) What is the length of AB .

b) Find cos C .

${\huge{\rm{\underline{\underline{Answer\checkmark}}}}}$

$\frak{Given}\begin{cases}\textsf{ABC\;is\;a\;right\;angled\;triangle.}\\ \textsf{Angle \;B = 90\degree }\\ \sf{AC=5cm}\\ \sf{sin\:C=\rm\dfrac{4}{5}}\end{cases}$

★ To Find :

• Length of AB
• cos C

★ Solution :

${\underline{\textit{\textbf{Finding \;AB}}}}$

$\rm \mapsto \: sin \: c = \frac{4}{5}$

$\mapsto \rm \: \frac{perpendicuar}{hypotenuse} = \frac{4}{5}$

$\mapsto\rm \; \dfrac{AB}{AC} =\dfrac{4}{5}$

$\mapsto\rm\; \dfrac{AB}{\cancel{5}}=\dfrac{4}{\cancel5}$

$\mapsto\underline{\boxed{\rm AB=4cm}}$

${\underline{\textit{\textbf{Finding\;cos\;c}}}}$

We have to first find the length of BC ..

Using Pythagoras Theorem -

AB² + BC² = AC²

4² + BC² = 5²

16 + BC² = 25

BC² = 25 - 16

BC² = 9

BC = 3 cm .

Now ,

$\rm \mapsto \: cos \: c = \frac{base}{hypotenuse}$

$\rm\mapsto \; cos\;c=\dfrac{BC}{AC}$

$\mapsto \rm \: cos \: c = \frac{3\: cm}{5 \: cm}$

$\mapsto \underline{\boxed{ \rm{cos \: c = \frac{3}{5}}} }$

So , Required answer :

1. AB = 4 cm
2. cos c = $\dfrac{3}{5}$