anonymoususer123
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In ΔPQR, sin R = [tex]\frac{3}{5}[/tex]. What is cos P?


(a) [tex]\frac{5}{3}[/tex]

(b) [tex]\frac{3}{4}[/tex]

(c) [tex]\frac{4}{3}[/tex]

(d) [tex]\frac{3}{5}[/tex]


(image attached below)

In ΔPQR sin R texfrac35tex What is cos Pa texfrac53texb texfrac34texc texfrac43texd texfrac35teximage attached below class=

Respuesta :

Answer:

Since we know that ΔPQR is a right triangle, we can also asume that:

sin R = cos P = 3/5

So the answer is (d).

* This formular can also be applied to other right triangles.

In a right triangle, sine of one acute angle will always be equal to cosine of the other acute angle.

And we can check this by actually finding cos P using the lengths of the sides, by calculating PR first:

PR = √(PQ² + RQ²) = √(12² + 16²) = 20

=> cos P = PQ/PR = 12/20 = 3/5