Question

In a right angled trìangle
length of base is 4 cm and
its perpendicular is 3 cm.
Find cos e​

Answer 1

Given

In Right Angled Triangle

⇒Base(b) = 4cm

⇒Perpendicular(p) = 3cm

To Find

⇒Cos(e)

We know that

⇒Cos(e) = Base(b)/Hypotenuse(h)

We have

⇒Base(b) = 4cm , Perpendicular(p) = 3cm and Hypotenuse(h) = x

Now Using Pythagoras theorem

⇒h² = p² + b²

Put the value

⇒h² = (3)² + (4)²

⇒h² = 9 + 16

⇒h² = 25

⇒h = 5cm

we get

⇒Base(b) = 4cm , Perpendicular(p) = 3cm and Hypotenuse(h) = 5cm

Then

⇒Cos(e) = Base(b)/Hypotenuse(h) = 4cm/5cm

Answer

⇒Cos(e) = 4cm/5cm

Answer 2

Answer:

Given :-

• In a right angled triangle length of base is 4 cm and it's perpendicular is 3 cm.

To Find :-

• What is the value of cose.

Formula Used :-

$\longmapsto \sf\boxed{\bold{\pink{{(Hypotenuse)}^{2} =\: {(Perpendicular)}^{2} + {(Base)}^{2}}}}$

$\longmapsto \sf\boxed{\bold{\pink{Cos\theta =\: \dfrac{Base}{Hypotenuse}}}}$

Solution :-

First, we have to find the value of Hypotenuse :

Given :

• Perpendicular = 3 cm
• Base = 4 cm

According to the question by using the formula we get,

$\implies \sf {(Hypotenuse)}^{2} =\: {(3)}^{2} + {(4)}^{2}$

$\implies \sf {(Hypotenuse)}^{2} =\: 3 \times 3 + 4 \times 4$

$\implies \sf {(Hypotenuse)}^{2} =\: 9 + 16$

$\implies \sf {(Hypotenuse)}^{2} =\: 25$

$\implies \sf Hypotenuse =\: \sqrt{25}$

$\implies \sf\bold{\green{Hypotenuse =\: 5\: cm}}$

Hence, the hypotenuse is 5 cm .

Now, we have to find the value of cose :

Given :

• Base = 4 cm
• Hypotenuse = 5 cm

According to the question by using the formula we get,

$\leadsto \sf Cose =\: \dfrac{Base}{Hypotenuse}$

$\implies \sf\bold{\red{Cose =\: \dfrac{4}{5}}}$

$\therefore$ The value of cose is 4/5.