Question

Give a perfect solution.​

Answer 1

$\huge\bold\blue{Solution:—}$

In ∆ acb, we have

$ac = \sqrt{a {b }^{2} } - b {c}^{2}$

$= \sqrt{(29 - 21)(29 + 21)}$

$= \sqrt{(8)(50)}$

$= \sqrt{400}$

$=20 \: units$

So,

$sin = \frac{ac}{ab}$

$= \frac{20}{29} \: \cos = \frac{bc}{ab} = \frac{21}{29}$

Now,

$(i) { \cos }^{2} + { \sin}^{2}$

$= {( \frac{20}{29}) }^{2} + {( \frac{21}{29}) }^{2}$

$= \frac{ {20}^{2} + {21}^{2} }{ {29}^{2} }$

$= \frac{400 + 441}{841} = 1$

$and \: (ii) { \cos }^{2} + { \sin }^{2} = (\frac{21}{29} {)}^{2} - ( \frac{20}{29} {)}^{2}$

$= \frac{(21 + 20)(21 - 20)}{ {29}^{2} }$

$= \frac{41}{841}$

Answer 2

Explanation:

Opp = Opposite

Hyp = Hypotenuse

Adj = Adjacent

Hope its clear