Question

In ΔABC, ∠A = 90° , AC=5 , BC=13 , the find sinB, tanC.​

Answer 1

➣ $\underline\bold\blue{Given}$

⭐∠A = 90°

⭐AC = 5

⭐BC = 13

➣ $\underline\bold\blue{To Find}$

⭐sin B = ?

⭐tan c = ?

➣ $\underline\bold\blue{solution}$

In Δ ABC ,

By Pythagoras Theorem

$\implies$BC² = AB² + AC²

$\implies$13². = AB + 5²

$\implies$ AB² = 169 - 25

$\implies$AB = $\sqrt{144}$

$\implies$AB = 12

Now we have

AB = 12cm

AC = 5cm

BC = 13 cm

________________________________

$\big\purple\star$ sin = $\dfrac{</strong><strong>perpendicular</strong><strong>}{</strong><strong>Hypo</strong><strong>tonuse</strong><strong>}$

$\big\purple\star$ tan = $\dfrac{perpendicular}{</strong><strong>Base</strong><strong>}$

________________________________

⟶ sinB = $\dfrac{5}{13}$

⟶ tanC = $\dfrac{12}{5}$

________________________________

Hope it helps

Answer 2

Answer:

➣ \underline\bold\blue{Given}

Given

⭐∠A = 90°

⭐AC = 5

⭐BC = 13

➣ \underline\bold\blue{To Find}

ToFind

⭐sin B = ?

⭐tan c = ?

➣ \underline\bold\blue{solution}

solution

In Δ ABC ,

By Pythagoras Theorem

\implies⟹ BC² = AB² + AC²

\implies⟹ 13². = AB + 5²

\implies⟹ AB² = 169 - 25

\implies⟹ AB = \sqrt{144}

144

\implies⟹ AB = 12

Now we have

AB = 12cm

AC = 5cm

BC = 13 cm

________________________________

\big\purple\star sin = \dfrac{perpendicular}{Hypotonuse}

Hypotonuse

perpendicular

\big\purple\star tan = \dfrac{perpendicular}{Base}

Base

perpendicular

________________________________

⟶ sinB = \dfrac{5}{13}

13

5

⟶ tanC = \dfrac{12}{5}

5

12

________________________________

Hope it helps