Alherreraidn
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pls help me with this!

Pls Help Me With This class=

Sagot :

Ivycoveredwallsidn

Answer:

0.69

Step-by-step explanation:

In a right triangle, the cosine of one of the acute angles is the ratio of the adjacent ("next to") side to the hypotenuse.

In this triangle, the side adjacent to angle T has length [tex]\sqrt{38}[/tex].  Now you need the length of the hypotenuse, ST.  Use the Pythagorean Theorem:

[tex]ST^2=(\sqrt{38})^2+(\sqrt{41})^2 \\\\ST^2=38+41\\\\ST=\sqrt{79}[/tex]

Build the ratio (adjacent) / (hypotenuse) and approximate it with a decimal.

[tex]\cos{T}=\frac{\sqrt{38}}{\sqrt{79}} \approx 0.69[/tex]

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To find :-

Cos T = ?

Given :-

Perpendicular(P) = RS = √41

Base(B) = RT = √38

Formula to be used :-

We will use here a trignometry formula to find hypotenuse (H) which is ST.

(hypotenuse)² = (perpendicular)² + (base)²

[tex] \cos(A) = \frac{base}{hypotenuse} [/tex]

Solution:-

H² = P² + B²

H² = (√41)² + (√38)²

H² = 41 + 38

H² = 79

H = √79

[tex] \cos(T) = \frac{B}{H} \\ \cos(T) = \frac{ \sqrt{38} }{ \sqrt{79} } \\ \cos(T) = \frac{6.17}{8.89} \\ \cos(T) = 0.69[/tex]

Result :-

The value of cos(T) is 0.69.

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