Question

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

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,

AC2 = AB2 + BC2

= 242  + 72

= 576 + 49

= 625 

∴ Hypotenuse AC = √625 cm = 25 cm

(i)

sin A = side opposite to ∠A / hypotenuse = BC/AC

sin A = 7 cm /25 cm = 7/25

sin A = 7/25

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

= 24 cm / 25 cm = 24/25

cos A = 24/25

(ii)

sin C = side opposite to ∠C / hypotenuse = AB/AC

sin C = 24 cm / 25 cm = 24/25

sin C = 24/25

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

= 7 cm / 25 cm = 7/25

cos C = 7/25

Answer 2

Answer:

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.

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,

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,AC2 = AB2 + BC2

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,AC2 = AB2 + BC2= 242  + 72

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,AC2 = AB2 + BC2= 242  + 72= 576 + 49

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,AC2 = AB2 + BC2= 242  + 72= 576 + 49= 625 

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,AC2 = AB2 + BC2= 242  + 72= 576 + 49= 625 ∴ Hypotenuse AC = √625 cm = 25 cm

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,AC2 = AB2 + BC2= 242  + 72= 576 + 49= 625 ∴ Hypotenuse AC = √625 cm = 25 cm(i)

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,AC2 = AB2 + BC2= 242  + 72= 576 + 49= 625 ∴ Hypotenuse AC = √625 cm = 25 cm(i)sin A = side opposite to ∠A / hypotenuse = BC/AC

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,AC2 = AB2 + BC2= 242  + 72= 576 + 49= 625 ∴ Hypotenuse AC = √625 cm = 25 cm(i)sin A = side opposite to ∠A / hypotenuse = BC/ACsin A = 7 cm /25 cm = 7/25

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,AC2 = AB2 + BC2= 242  + 72= 576 + 49= 625 ∴ Hypotenuse AC = √625 cm = 25 cm(i)sin A = side opposite to ∠A / hypotenuse = BC/ACsin A = 7 cm /25 cm = 7/25sin A = 7/25

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,AC2 = AB2 + BC2= 242  + 72= 576 + 49= 625 ∴ Hypotenuse AC = √625 cm = 25 cm(i)sin A = side opposite to ∠A / hypotenuse = BC/ACsin A = 7 cm /25 cm = 7/25sin A = 7/25cos A = side adjacent to ∠A / hypotenuse = AB/AC

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,AC2 = AB2 + BC2= 242  + 72= 576 + 49= 625 ∴ Hypotenuse AC = √625 cm = 25 cm(i)sin A = side opposite to ∠A / hypotenuse = BC/ACsin A = 7 cm /25 cm = 7/25sin A = 7/25cos A = side adjacent to ∠A / hypotenuse = AB/AC= 24 cm / 25 cm = 24/25

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,AC2 = AB2 + BC2= 242  + 72= 576 + 49= 625 ∴ Hypotenuse AC = √625 cm = 25 cm(i)sin A = side opposite to ∠A / hypotenuse = BC/ACsin A = 7 cm /25 cm = 7/25sin A = 7/25cos A = side adjacent to ∠A / hypotenuse = AB/AC= 24 cm / 25 cm = 24/25cos A = 24/25

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,AC2 = AB2 + BC2= 242  + 72= 576 + 49= 625 ∴ Hypotenuse AC = √625 cm = 25 cm(i)sin A = side opposite to ∠A / hypotenuse = BC/ACsin A = 7 cm /25 cm = 7/25sin A = 7/25cos A = side adjacent to ∠A / hypotenuse = AB/AC= 24 cm / 25 cm = 24/25cos A = 24/25(ii) sin C = side opposite to ∠C / hypotenuse = AB/AC

C = side opposite to ∠C / hypotenuse = AB/ACsin C = 24 cm / 25 cm = 24/25

C = side opposite to ∠C / hypotenuse = AB/ACsin C = 24 cm / 25 cm = 24/25sin C = 24/25

C = side opposite to ∠C / hypotenuse = AB/ACsin C = 24 cm / 25 cm = 24/25sin C = 24/25cos C = side adjacent to ∠C / hypotenuse = BC/AC

= "7 cm / 25 cm = 7/25cos C = 7/25."

"Hope this Helpful."