Answer:
[tex]\textsf{4)} \quad \text{D.}\;\; \dfrac{\sqrt{3}}{3}[/tex]
5) C. Equal
Step-by-step explanation:
Question 4
The cotangent of an angle in a right triangle is the reciprocal of the tangent of that angle:
[tex]\cot\theta=\dfrac{1}{\tan\theta}[/tex]
Given that tan 60° = √3, then:
[tex]\cot 60^{\circ}=\dfrac{1}{\tan 60^{\circ}}\\\\\\\cot 60^{\circ}=\dfrac{1}{\sqrt{3}}\\\\\\\cot 60^{\circ}=\dfrac{1\cdot \sqrt{3}}{\sqrt{3}\cdot \sqrt{3}}\\\\\\\cot 60^{\circ}=\dfrac{\sqrt{3}}{3}[/tex]
Therefore, the value of cot 60° is:
[tex]\Large\boxed{\boxed{\dfrac{\sqrt{3}}{3}}}[/tex]
[tex]\dotfill[/tex]
Question 5
If ∠M = 90° and ∠I = 30° in triangle KIM, then according to the Angle Sum Property of a Triangle, ∠K = 60°. Since one of the angles it a right angle, triangle KIM is a right triangle.
In a right triangle, the sine of an angle is defined as the ratio of the length of the side opposite the angle to the length of the hypotenuse, and the cosine of an angle is defined as the ratio of the length of the side adjacent to the angle to the length of the hypotenuse.
[tex]\boxed{\begin{array}{l}\underline{\sf Trigonometric\;ratios}\\\\\sf \sin(\theta)=\dfrac{O}{H}\qquad\cos(\theta)=\dfrac{A}{H}\\\\\textsf{where:}\\\phantom{ww}\bullet\;\textsf{$\theta$ is the angle.}\\\phantom{ww}\bullet\;\textsf{O is the side opposite the angle.}\\\phantom{ww}\bullet\;\textsf{A is the side adjacent the angle.}\\\phantom{ww}\bullet\;\textsf{H is the hypotenuse (the side opposite the right angle).}\end{array}}[/tex]
As angle M in ΔKIM is the right angle, side KI is the hypotenuse.
As ∠I = 30°, then KM is the side opposite ∠I, and IM is the side adjacent to ∠I. Therefore:
[tex]\sin 30^{\circ}=\sin I=\dfrac{KM}{KI}[/tex]
As ∠K = 60°, then IM is the side opposite ∠K, and KM is the side adjacent to ∠K. Therefore:
[tex]\cos 60^{\circ}=\cos K=\dfrac{KM}{KI}[/tex]
So, the relation that exists between sin 30° and cos 60° is that they are:
[tex]\Large\boxed{\boxed{\sf Equal}}[/tex]
