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find: \\sin a\ version 1: pythagorean triples only

Question

find: \sin a\
version 1: pythagorean triples only

Explanation:

Step1: Identify triangle sides

In right triangle \(ABC\) (right - angled at \(B\)), hypotenuse \(AC = 25\), \(AB=20\). First, find the length of \(BC\) using the Pythagorean theorem \(AC^{2}=AB^{2}+BC^{2}\). So, \(BC=\sqrt{AC^{2}-AB^{2}}=\sqrt{25^{2}-20^{2}}=\sqrt{(25 + 20)(25 - 20)}=\sqrt{45\times5}=\sqrt{225}=15\).

Step2: Recall sine definition

The sine of an angle in a right triangle is defined as the ratio of the length of the opposite side to the length of the hypotenuse. For \(\angle A\), the opposite side is \(BC\) and the hypotenuse is \(AC\). So, \(\sin A=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{BC}{AC}\).

Step3: Substitute values

We know that \(BC = 15\) and \(AC=25\). So, \(\sin A=\frac{15}{25}=\frac{3}{5}\).

Answer:

\(\frac{3}{5}\)