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laiz [17]
3 years ago
5

How do I solve for what is sin A for the triangle shown

Mathematics
1 answer:
PtichkaEL [24]3 years ago
7 0

Option a:

sin A for the triangle is \frac{3}{5}.

Solution:

The given triangle is a right triangle.

The adjacent side to angle A is AC.

The opposite side to angle A is BC.

Hypotenuse is AB.

AC = 8, BC = 6 and AB = 10.

Using trigonometric ratio formulas,

$\sin \theta=\frac{\text { Opposite side }}{\text { Hypotenuse }}

$\sin A=\frac{BC}{AB}

$\sin A=\frac{6}{10}

Divide both numerator and denominator by 2, we get

$\sin A=\frac{6\div2}{10\div 2}

$\sin A=\frac{3}{5}

Hence sin A for the triangle is \frac{3}{5}.

Option a is the correct answer.

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