Question

 In ∆ ABC, right-angled at B, AB = 24 cm, BC = 7 cm.​

Answer 1

Answer:-

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

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Given:-

• AB = 24 cm

• BC = 7 cm

According to the Pythagoras Theorem, in a right-angled triangle, the squares of the hypotenuse side is equal to the sum of the squares of the other two sides.

By applying the Pythagoras theorem, we get

$AC {}^{2} = AB {}^{2} + BC {}^{2}$

$AC {}^{2} = ( {24})^{2} + {7}^{2}$

$AC {}^{2} = (576 + 49)$

$AC {}^{2} = 625cm {}^{2}$

$AC = \sqrt{625}$

$AC = 25$

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To find Sin (A), Cos (A)

We know that the sine (or) Sin function is equal to the ratio of the length of the opposite side to the hypotenuse side. So it becomes

Sin (A) = Opposite side /Hypotenuse

Sin (A) = BC/AC

Sin (A) = 7/25

Cosine or Cos function is equal to the ratio of the length of the adjacent side to the hypotenuse side and it becomes,

Cos (A) = Adjacent side/Hypotenuse

Cos (A) = AB/AC

Cos (A) = 24/25

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(ii) To find Sin (C), Cos (C)

Sin (C) = AB/AC

Sin (C) = 24/25

Cos (C) = BC/AC

Cos (C) = 7/25

Answer 2

Topic:- Triangles,Pythagorean theorem,Euclids Geometry,Plane Geometry

Answer:

$\sf \: AC = 25cm$

Step-by-step explanation:

As given in attachement.

AB = 24cm

BC = 7cm

AC = x

Let's write Hypotenuse,Sides

HYPOTENUSE = AC = x

Side = AB = 24cm

Side = BC = 7cm

We know that:-

$\sf \boxed { \sf \:Hypotenuse² = (Side)²+(Side)²}$

By that formula :-

$\sf \: AC²=AB²+BC²$

Putting the values:-

$\sf \: (x)² = (24)²+(7)²$

$\sf \implies \: (x)² = 576 + 49$

$\sf \implies \: {(x)}^{2} = 625$

$\sf \implies \: x = \sqrt{625} = 25$

x = AC = 25