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Answer:

[tex] \star \sf {\boxed{ \red{ \sf hello \: pooja}}}[/tex]

Given that :-

• AB=24 cm
• BC=4 cm

To find

• Determining

1. sin A, COS A
2. sin C, cos C

Formula used:-

• We use the basic formulas of trigonometric ratios to solve the question.

Quick tip:-

• According to the Pythagoras Theorem, in a right-angled triangle, the squares of the hypotenuse side is equal to the sum of the squares of the other two sides.

Solution:-

[tex] \sf \: AC ^{2} = {AB}^{2} + {BC}^{2} \\ \\ \sf {AC}^{2} = {(24)}^{2} + {(7)}^{2} \\ \\ \sf \: {AC}^{2} = (576 + 49) \\ \\ \sf {AC}^{2} = 625 {cm}^{2} \\ \\ \therefore \sf \: AC = \sqrt{625} \: \: ( \sqrt{25 \times 25} ) \\ \\ \sf \: AC = 25cm[/tex]

i) sin A, COS A

We know that the sine (or) Sin function is equal to the ratio of the length of the opposite side to the hypotenuse side.

[tex] \sf \: \sin(A) = \frac{opposite \: side}{hypotenuse} \\ \\ \sf \: \sin(A) = \frac{7}{25} [/tex]

Cosine or Cos function is equal to the ratio of the length of the adjacent side to the hypotenuse side and it becomes

[tex] \sf \: \cos(A) = \frac{adjacent \: side}{hypotenuse} \\ \\ \sf \cos(A) = \frac{24}{25} [/tex]

ii)sin C, cos C

[tex] \sf \sin(C) = \frac{AB}{AC} \\ \\ \sf \sin(C) = \frac{24}{25} \\ \\ \sf \cos(C) = \frac{BC}{AC} \\ \\ \sf \cos(C) = \frac{7}{25} [/tex]

Hope it helps you from my side!!

Keep Learning!!

[tex] \rule{300pt}{2.5pt}[/tex]

Solution :-

• We use the basic formulas of trigonometric ratios to solve the question.

⚽ Applying the Pythagoras theorem for ∆ABC, we can find the hypotenuse (side AC). Once hypotenuse is known, we can find sine and cosine angles using trigonometric ratios.

➢ In ΔABC, we obtain,

[tex] \begin{gathered}\\ \large\implies\sf{ AC2 = AB2 + BC2 } \\ \end{gathered}[/tex]

[tex] \begin{gathered}\\ \large\implies\sf{ 242 + 72 } \\ \end{gathered}[/tex]

[tex] \begin{gathered}\\ \large\implies\sf{ 576 + 49 } \\ \end{gathered}[/tex]

[tex] \begin{gathered}\\ \large\implies\bf{ 625 } \\ \end{gathered}[/tex]

∴ Hypotenuse AC = √625 cm = 25 cm sin A = side opposite to ∠A / hypotenuse = BC/AC

• ➢ sin A = [tex] \sf{\frac{7 cm }{25cm}}[/tex]

• ➢ sin A = [tex] \sf{\frac{7}{25}}[/tex]

→ cos A = side adjacent to ∠A / hypotenuse = AB/AC

[tex] \sf{\frac{24cm}{25cm}}[/tex]= [tex] \sf{\frac{24}{25}}[/tex]

→ cos A = 24/25sin C = side opposite to ∠C / hypotenuse = AB/AC

➢ sin C = [tex] \sf{\frac{24cm}{25cm}}[/tex]= [tex] \sf{\frac{24}{25}}[/tex]

➢ sin C = [tex] \sf{\frac{24}{25}}[/tex]

→ cos C = side adjacent to ∠C / hypotenuse = BC/AC

[tex] \sf{\frac{7cm}{25cm}}[/tex] = [tex] \sf{\frac{7}{25}}[/tex]

➢ cos C = [tex] {\large\bf\red{\frac{7}{25}}}[/tex]

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