Respuesta :
Answer:
Option B is correct.
Value of cos(a+B) is, [tex]\frac{16}{65}[/tex]
Step-by-step explanation:
Using the formula:
[tex]\cos(a+B) = \cos a \cos B -\sin a \sin B[/tex]
Given the values:
[tex]\cos a = \frac{3}{5}[/tex] and [tex]\sin B = \frac{5}{13}[/tex]
Using trigonometric identity:
[tex]\sin^2 \theta + \cos^2 \theta =1[/tex]
or
[tex]\sin \theta = \sqrt{1-\cos^2 \theta}[/tex]
or
[tex]\cos \theta = \sqrt{1-\sin^2 \theta}[/tex]
Find the value of sin a and cos B;
Using trigonometric identity:
[tex]\sin a = \sqrt{1-\cos^2 a}[/tex]
[tex]\sin a = \sqrt{1-(\frac{3}{5})^2}[/tex] = [tex]\sqrt{1-\frac{9}{25}} = \sqrt{\frac{25-9}{25}}=\sqrt{\frac{16}{25} } =\frac{4}{5}[/tex]
Similarly;
[tex]\cos B = \sqrt{1-\sin^2 B}[/tex]
[tex]\cos B = \sqrt{1-(\frac{5}{13})^2}[/tex] = [tex]\sqrt{1-\frac{25}{169}} = \sqrt{\frac{169-25}{169}}=\sqrt{\frac{144}{169} } =\frac{12}{13}[/tex]
Substitute these given values equation [1] we have;
[tex]\cos(a+B) =\frac{3}{5} \cdot \frac{12}{13}-\frac{4}{5} \cdot \frac{5}{13}[/tex]
[tex]\cos(a+B) =\frac{36}{65}-\frac{20}{65}[/tex]
then;
[tex]\cos(a+B) =\frac{36-20}{65}=\frac{16}{65}[/tex]
Since, the angles a and B are in first quadrant angles therefore the value of cos(a+B) is, [tex]\frac{16}{65}[/tex]
