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baabaagreysheep
sin D - 4/5
This is because sin would mean Opposite/Hypotenuse, and so if the angle is D then the opposite is 4.

sin C - 3/5
This is basically the same reason as sin D, however this time the Opposite is 3.

sin D × cos D - 4/5 × 3/5 = 12/25
For here, you just get the fractions of sin D, which we got earlier, and cos D, which would be Adjacent/Hypotenuse and therefore 3/5.

tan C × tan D - Not an option

cos C × tan D - 16/15
This is because cos C would be 4/5, and tan D would be 4/3 (Opposite/Adjacent) and multiplying those fractions together gives 16/15.

I hope this helps!

You can use the definitions of first three trigonometric ratios to find out the correct boxes.

The correct boxes filled would be like this

[tex]\rm \dfrac{12}{25} = \cos C \times \cos D = \sin C \times \sin D\\\\\dfrac{3}{5} = \sin C\\\\\dfrac{16}{25} = \cos C \times \tan D\\\\\dfrac{4}{5} = \cos C = \sin D[/tex]

What are the first three trigonometric ratios?

They are sine (sin), cosine (cos), tangent (tan)

They are ratios of sides of a right angled triangle, viewed from a fixed angle (not 90 degrees but the rest of the two angles)

Suppose we have the right angled triangle (as in given figure) CBD when angle B is right angle(of 90 degrees), then let the angle we choose be angle C.

Then we have:

The hypotenuse is always same, the slant line = CD

The side which is opposite to angle C is the line BD, called perpendicular from the viewpoint of angle C

And the line which is remaining now is called the base = CB

Thus, for these lines, we have the ratios as

[tex]sin(\angle C) = \dfrac{\text{Perpendicular}}{\text{Hypotenuse}} = \dfrac{|BD|}{|CD|} = \dfrac{3}{5}\\\\cos(\angle C) = \dfrac{\text{Base}}{\text{Hypotenuse}} = \dfrac{|CB|}{|CD|} = \dfrac{4}{5}\\\\\\tan(\angle C) = \dfrac{\text{Perpendicular}}{\text{Base}} = \dfrac{|BD|}{|CB|} = \dfrac{3}{4}\\[/tex]

Using the above facts, we get the ratios as

From angle C:

[tex]sin(\angle C) = \dfrac{\text{Perpendicular}}{\text{Hypotenuse}} = \dfrac{|BD|}{|CD|} = \dfrac{3}{5}\\\\cos(\angle C) = \dfrac{\text{Base}}{\text{Hypotenuse}} = \dfrac{|CB|}{|CD|} = \dfrac{4}{5}\\\\\\tan(\angle C) = \dfrac{\text{Perpendicular}}{\text{Base}} = \dfrac{|BD|}{|CB|} = \dfrac{3}{4}\\[/tex]

From angle D:

[tex]sin(\angle D) = \dfrac{\text{Perpendicular}}{\text{Hypotenuse}} = \dfrac{|CB|}{|CD|} = \dfrac{4}{5}\\\\cos(\angle D) = \dfrac{\text{Base}}{\text{Hypotenuse}} = \dfrac{|BD|}{|CD|} = \dfrac{3}{5}\\\\\\tan(\angle D) = \dfrac{\text{Perpendicular}}{\text{Base}} = \dfrac{|CB|}{|BD|} = \dfrac{4}{3}\\[/tex]

Thus,

[tex]sinC \times cosD = 3/5 \times 3/5 = 9/25[/tex]
[tex]tanC \times tanD = 4/3 \times 3/4 = 1[/tex]

[tex]cosC \times tanD = 4/5 \times 4/3 = 16/15[/tex]

Thus,

[tex]\rm \dfrac{12}{25} = \cos C \times \cos D = \sin C \times \sin D\\\\\dfrac{3}{5} = \sin C\\\\\dfrac{16}{25} = \cos C \times \tan D\\\\\dfrac{4}{5} = \cos C = \sin D[/tex]

Learn more about trigonometric ratios here;

https://brainly.com/question/22599614

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