Question

In a triangle ABC right angled at B, AB = 5 cm and AC = 10 cm. Determine ∠ BAC and ∠BCA.
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Answer 1

$In \:right \: triangle \: ABC , \angle B = 90\degree$

$and \: AB = 5 \:cm , \: AC = 10 \:cm$

$i ) Sin C = \frac{ AB}{AC}$

$\implies Sin C = \frac{5}{10}$

$\implies Sin C = \frac{1}{2}$

$\implies Sin C = Sin 30 \degree$

$\implies C = 30 \degree$

$ii) Cos A = \frac{ AB}{AC}$

$\implies Cos A = \frac{5}{10}$

$\implies Cos A = \frac{1}{2}$

$\implies Cos A = Cos 60 \degree$

$\implies A = 60 \degree$

Therefore.,

$\green { \angle A = 60\degree \:and \:\angle C = 30\degree}$

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Answer 2

Answer:

In ΔABC, <B = 90°

= AB = 5cm

= AC = 10cm

= Sin θ = 5/10 = 1/2

= <BCA = $sin^{-1}$ (1/2) = 30°

= <BCA = 30°

= <BAC = 90° - <BCA (30°)

= <BAC = 90° - 30° = 60°

= <BAC = 60°

∴ <BCA = 30° and <BAC = 60°