Help me with a explanation please!!

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Mathematics

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Help me with a explanation please!!

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Semsee45id Semsee45id · Answer 1

Answer:

[tex]\sf sec(D)=\sqrt{2}[/tex]

Step-by-step explanation:

First, find the length of the hypotenuse using Pythagoras' Theorem:

[tex]a^2+b^2=c^2[/tex]

(where a and b are the legs, and c is the hypotenuse, of a right triangle)

Given:

[tex]a=4\sqrt{10}[/tex]
[tex]b=4\sqrt{10}[/tex]

Subtituting the given values into the formula and solving for c:

[tex]\implies (4\sqrt{10})^2+(4\sqrt{10})^2=c^2[/tex]

[tex]\implies 160+160=c^2[/tex]

[tex]\implies c^2=320[/tex]

[tex]\implies c=\sqrt{320}[/tex]

[tex]\implies c=8\sqrt{5}[/tex]

Therefore, the hypotenuse is [tex]\sf 8\sqrt{5}[/tex]

Secant of an angle in a right triangle:

[tex]\sf \sec(\theta)=\dfrac{\sf hypotenuse}{\sf adjacent\:side}[/tex]

Given:

[tex]\theta[/tex] = D
hypotenuse = [tex]\sf 8\sqrt{5}[/tex]
adjacent side = [tex]\sf 4\sqrt{10}[/tex]

Substituting the given values into the formula:

[tex]\implies \sf \sec(D)=\dfrac{8\sqrt{5}}{4\sqrt{10}}=\sqrt2}[/tex]

Анонимid Анонимid · Answer 2

Answer:

[tex] \sf{\Large{\bold{\orange{The \: value \: of \: \sec (D) \: is \: \sqrt{2} }}}}[/tex]

Step-by-step explanation:

[tex] \textsf{\large{\underline{\green{To find:-}}}}[/tex]

The value of Sec(D)

[tex] \textsf{\large{\underline{\blue{Given :-}}}}[/tex]

Perpendicular (P) = BC = 4√10

Base (B) = BD = 4√10

[tex] \textsf{\huge{\underline{\underline{\pink{Solution :-}}}}}[/tex]

First of all we have to find length of DC to get hypotenuse

To find we will use Pythagoras theorem

[tex] \sf {(hypotenuse)}^{2} = {(perpendicular)}^{2} + {(base)}^{2} \\ \sf {H}^{2} = {P}^{2} + {B}^{2} [/tex]

Now we will put the values of Perpendicular and base

[tex]\sf {H}^{2} = {(4 \sqrt{10} )}^{2} + {(4 \sqrt{10} )}^{2} \\ \sf {H}^{2} = (16 \times 10) + (16 \times 10) \\ \sf {H}^{2} = 160 + 160 \\ \sf {H}^{2} = 320 \\ \sf H = \sqrt{320} \\ \sf H = 8 \sqrt{5}[/tex]

Hypotenuse (H) = CD = 8√5

Now we will put Sec formula

to find Sec(D)

[tex] \sf \sec(D) = \frac{hypotenuse}{base} = \frac{H}{B} \\ \implies \frac{8 \sqrt{5} }{4 \sqrt{10} } \\ \implies \frac{2 \sqrt{5} }{ \sqrt{10} } \\ \implies \frac{2 \sqrt{5} }{ \sqrt{2} \sqrt{5} } \\ \implies \frac{2}{ \sqrt{2} } \\ \implies{ \sf {\purple {\sqrt{2} }}}[/tex]