Question

Question. 1 : In ∆ ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine: (i) sin A, cos A (ii) sin C, cos C

Answer 1

ANSWER:

• (i) sin A = 7/25 , cos A = 24/25

• (ii) sin C = 24/25 , cos C = 7/25

GIVEN:

• Right-angled at B

• AB = 24 cm

• BC = 7 cm

TO FIND:

• (i) sin A , cos A

• (ii) sin C , cos C

EXPLANATION:

By using Pythagoras theorem:

The square of the hypotenuse will be equal to the sum of the squares of the other two sides.

AC² = BC² + AB²

AC² = 24² + 7²

AC² = 576 + 49

AC² = 625

AC = 25 cm

$\rule{200}{1}$

(i) Sin A = opposite side / hypotenuse

BC is the opposite side of ∠A

Sin A = 7  /  25

Cos A = adjacent side / hypotenuse

AB is the Adjacent  side of ∠A

Cos A = 24  /  25

$\rule{200}{1}$

(ii) Sin C = opposite side / hypotenuse

AB is the opposite  side of ∠C

Sin C = 24  /  25

Cos C = adjacent / hypotenuse

BC is the adjacent  side of ∠C

Cos C = 7  /  25

REFER ATTACHMENT FOR DIAGRAM.

Answer 2

$\tt \huge \underline\pink{question}$

In ∆ ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine: (i) sin A, cos A (ii) sin C, cos C

$\tt \huge \underline\pink{answer}$

$\bf \green{ \: in \: right \: angled \: \triangle \: ABC \: we \: have \: AB = 24 \: cm \: and \: BC = 7 \: cm}$

$\bf \red{ \therefore \: using \: pythagoras \: theorem \: {AC}^{2} = {AB}^{2} + {BC}^{2} }$

$\bf \bold{ \implies \: {AC}^{2} = {24}^{2} + {7}^{2} = 576 + 49 = 625 = {25}^{2} }$

$\bf \bold{ \implies \: AC = 25 \: cm}$

$\bf \: 1) \purple{sin \: A = \frac{BC}{AC} = \frac{7}{25} } \: \: \bf{and} \: \: \: \bf \purple{cos \: A = \frac{AB}{AC} = \frac{24}{25}}$

$\bf 2) \purple{ \: sin \: C = \frac{AB}{AC} = \frac{24}{25} } \: \: \: \bf \:{ and} \: \: \: \bf \purple{ \: cos \: C = \frac{BC}{AC} = \frac{7}{25} }$

TRIGONOMETRIC ratios :-

$\bf \huge \star{ \: sin \theta \: = \frac{P}{H} }$

$\bf \huge \star{cos \: \theta = \frac{B}{H} }$

$\bf \huge \star{tan \: \theta = \frac{P}{B} }$