Question

In triangle DEF, right angled at e, de = 4 cm and fe = 3 cm. find the values of sin D, sin F, sec D and sec F. ​

Answer 1

S O L U T I O N :

$\underline{\bf{Given\::}}$

In Δ DEF, right angled at E, DE = 4 cm & FE = 3 cm.

$\underline{\bf{Explanation\::}}$

Firstly, attachment a figure of right angled Δ according to the given question.

We have two sides;

• DE = 4 cm
• FE = 3 cm

$\underline{\underline{\tt{Using\:\:by\:\:Pythagoras\:\:theorem\::}}}$

$\mapsto\sf{(Hypotenuse)^{2} = (Base)^{2} + (Perpendicular)^{2} }$

$\mapsto\sf{(FD)^{2} = (FE)^{2} + (DE)^{2} }$

$\mapsto\sf{(FD)^{2} = (3\:cm)^{2} + (4\:cm)^{2} }$

$\mapsto\sf{(FD)^{2} = 9\:cm^{2} + 16\:cm^{2} }$

$\mapsto\sf{(FD)^{2} = 25\:cm^{2} }$

$\mapsto\sf{FD = \sqrt{25\:cm^{2}} }$

$\mapsto\bf{FD = 5\:cm }$

Now, the values get :

As we know that,

$\boxed{\bf{sin\:\theta = \frac{Perpendicular }{Hypotenuse} }}$

$\mapsto\tt{sin\:D = \dfrac{FE}{FD}}$

$\mapsto\bf{sin\:D = \dfrac{3\:cm}{5\:cm}}$

&

$\mapsto\tt{sin\:F = \dfrac{DE}{FD}}$

$\mapsto\bf{sin\:F= \dfrac{4\:cm}{5\:cm}}$

As we know that,

$\boxed{\bf{sec\:\theta = \frac{Hypotenuse }{Base} }}$

Therefore,

$\mapsto\tt{sec\:D = \dfrac{FD}{DE}}$

$\mapsto\bf{sec\:D = \dfrac{5\:cm}{4\:cm}}$

&

$\mapsto\tt{sec\:F = \dfrac{FD}{FE}}$

$\mapsto\bf{sec\:F = \dfrac{5\:cm}{3\:cm}}$

Answer 2

Step-by-step explanation:

sin D = 3cm/5cm

sin F = 4cm / 5 cm

sec D = 5cm / 4 cm