Question

In a triangle ABC right angled at B, AB = 24 cm, BC = 7 cm. then sinC = ?​

Answer 1

Answer:

In Δ ABC, right-angled at B

Using Pythagoras theorem

AC² = AB² +BC²

AC² = 576 + 49 = 625

AC = √635

AC = +25

Now

AC = 25 CM, AB = 24cm , BC = 7cm

sinC=

side opposite to angle c / hypotenuse =>

AB / AC => 24/25

Answer 2

REQUIRED ANSWER : -

$\longrightarrow$ In ∆ ABC, ∠B = 90°

$\longrightarrow$ AB = 24 cm BC = 7 cm

( by pythyorous therom )

We have ,

➠$ac {}^{2} = ab {}^{2} + \: bc {}^{2}$

➠$ac = \: \sqrt{(24) {}^{2} } + (7) {}^{2}$

➠ $\sqrt{576 + 49}$

➠ $\sqrt{625}$

➠ AC = 25 cm

Now , sin A = $\frac{bc}{ac} = \frac{7}{25}$

Cos A = $\frac{ab}{ac} = \frac{24}{25}$

ANOTHER METHOD IS : -

➠ In a given triangle ABC, right-angled at B = ∠B = 90°

➠ Given: AB = 24 cm and BC = 7 cm

➠ That means, AC = Hypotenuse

➠ According to the Pythagoras Theorem,

➠In a right-angled triangle, the squares of the hypotenuse side are equal to the sum of the squares of the other two sides.

➠ By applying Pythagoras theorem, we get

➠ AC2 = AB2 + BC2

AC2 = (24)2 + 72

➠ AC2 = (576 + 49)

➠ AC2 = 625 cm2

➠ Therefore, AC = 25 cm

(i) We need to find Sin A and Cos A.

➠ As we know, sine of the angle is equal to the ratio of the length of the opposite side and hypotenuse side. Therefore,

➠ Sin A = BC/AC = 7/25

➠Again, the cosine of an angle is equal to the ratio of the adjacent side and hypotenuse side. Therefore,

➠ cos A = AB/AC = 24/25

(ii) We need to find Sin C and Cos C.

➠ Sin C = AB/AC = 24/25

➠Cos C = BC/AC = 7/25

hence proved ,