Answer :

[tex]A = 40, c = 41, b = ? \\ \\ \cos{\angle A}= \frac{c}{b} \\ \\ \cos{40^o}= \frac{41}{b} \\ \\b=\frac{41} {cos{40^o}}= 53.5[/tex]
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Answer:

Option C is correct

the value of b approximately is, 57 units

Step-by-step explanation:

Using Pythagoras theorem in a right angle triangle:

[tex]\text{Hypotenuse side}^2=\text{Opposite side}^2+\text{Adjacent side}^2[/tex]

As per the statement:

In ACB,

a = 40 units , c = 41

Solve for b:

Using Pythagoras theorem;

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

Substitute the given values we have;

[tex]b^2 = 40^2+41^2[/tex]

⇒[tex]b^2 = 3281[/tex]

⇒[tex]b = \sqrt{3281} = 57.280014[/tex] units

Therefore, the value of b approximately is, 57 units

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