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In triangle JKL, tan(b°) = three fourths and cos(b°) =four fifths. If triangle JKL is dilated by a scale factor of one half, what is sin(b°)?

triangle KL in which angle K is a right angle and angle L measures b degrees

sin(b°) = three fifths
sin(b°) = four fifths
sin(b°) = five thirds
sin(b°) = five fourths

1 answer

To find \(\sin(b^\circ)\) given \(\tan(b^\circ) = \frac{3}{4}\) and \(\cos(b^\circ) = \frac{4}{5}\), we can use the Pythagorean identity.

The Pythagorean identity states that: \[ \sin^2(b) + \cos^2(b) = 1 \]

Given that \(\cos(b) = \frac{4}{5}\), we can calculate \(\sin(b)\):

  1. Calculate \(\cos^2(b)\): \[ \cos^2(b) = \left(\frac{4}{5}\right)^2 = \frac{16}{25} \]

  2. Use the identity to find \(\sin^2(b)\): \[ \sin^2(b) = 1 - \cos^2(b) = 1 - \frac{16}{25} = \frac{25}{25} - \frac{16}{25} = \frac{9}{25} \]

  3. Now, take the square root to find \(\sin(b)\): \[ \sin(b) = \sqrt{\frac{9}{25}} = \frac{3}{5} \]

Therefore, \(\sin(b^\circ) = \frac{3}{5}\).

Thus, the correct answer is:

\(\sin(b^\circ) = \frac{3}{5}\)