Find the values of the sine, cosine, and tangent for ∠A
a. sin A = [tex]\frac{\sqrt{13} }{2}[/tex], cos A = [tex]\frac{\sqrt{13} }{3}[/tex], tan A = [tex]\frac{2 }{3}[/tex]
b. sin A = [tex]3\frac{\sqrt{13} }{13}[/tex], cos A = [tex]2\frac{\sqrt{13} }{13}[/tex], tan A = [tex]\frac{3}{2}[/tex]
c. sin A = [tex]\frac{\sqrt{13} }{3}[/tex], cos A = [tex]\frac{\sqrt{13} }{2}[/tex], tan A = [tex]\frac{3}{2}[/tex]
d. sin A = [tex]2\frac{\sqrt{13} }{13}[/tex], cos A = [tex]3\frac{\sqrt{13} }{13}[/tex], tan A = [tex]\frac{2 }{3}[/tex]
d. sin A = [tex]2\frac{\sqrt{13} }{13}[/tex], cos A = [tex]3\frac{\sqrt{13} }{13}[/tex], tan A = [tex]\frac{2 }{3}[/tex]
The triangle for the question has been attached to this response.
As shown in the triangle;
AC = 36ft
BC = 24ft
ACB = 90°
To calculate the values of the sine, cosine, and tangent of ∠A;
i. First calculate the value of the missing side AB.
Using Pythagoras' theorem;
⇒ (AB)² = (AC)² + (BC)²
Substitute the values of AC and BC
⇒ (AB)² = (36)² + (24)²
Solve for AB
⇒ (AB)² = 1296 + 576
⇒ (AB)² = 1872
⇒ AB = [tex]\sqrt{1872}[/tex]
⇒ AB = [tex]12\sqrt{13}[/tex] ft
From the values of the sides, it can be noted that the side AB is the hypotenuse of the triangle since that is the longest side with a value of [tex]12\sqrt{13}[/tex] ft (43.27ft).
ii. Calculate the sine of ∠A (i.e sin A)
The sine of an angle (Ф) in a triangle is given by the ratio of the opposite side to that angle to the hypotenuse side of the triangle. i.e
sin Ф = [tex]\frac{opposite}{hypotenuse}[/tex] -------------(i)
In this case,
Ф = A
opposite = 24ft (This is the opposite side to angle A)
hypotenuse = [tex]12\sqrt{13}[/tex] ft (This is the longest side of the triangle)
Substitute these values into equation (i) as follows;
sin A = [tex]\frac{24}{12\sqrt{13} }[/tex]
sin A = [tex]\frac{2}{\sqrt{13}}[/tex]
Rationalize the result by multiplying both the numerator and denominator by [tex]\sqrt{13}[/tex]
sin A = [tex]\frac{2}{\sqrt{13}} * \frac{\sqrt{13} }{\sqrt{13} }[/tex]
sin A = [tex]\frac{2\sqrt{13} }{13}[/tex]
iii. Calculate the cosine of ∠A (i.e cos A)
The cosine of an angle (Ф) in a triangle is given by the ratio of the adjacent side to that angle to the hypotenuse side of the triangle. i.e
cos Ф = [tex]\frac{adjacent}{hypotenuse}[/tex] -------------(ii)
In this case,
Ф = A
adjacent = 36ft (This is the adjecent side to angle A)
hypotenuse = [tex]12\sqrt{13}[/tex] ft (This is the longest side of the triangle)
Substitute these values into equation (ii) as follows;
cos A = [tex]\frac{36}{12\sqrt{13} }[/tex]
cos A = [tex]\frac{3}{\sqrt{13}}[/tex]
Rationalize the result by multiplying both the numerator and denominator by [tex]\sqrt{13}[/tex]
cos A = [tex]\frac{3}{\sqrt{13}} * \frac{\sqrt{13} }{\sqrt{13} }[/tex]
cos A = [tex]\frac{3\sqrt{13} }{13}[/tex]
iii. Calculate the tangent of ∠A (i.e tan A)
The cosine of an angle (Ф) in a triangle is given by the ratio of the opposite side to that angle to the adjacent side of the triangle. i.e
tan Ф = [tex]\frac{opposite}{adjacent}[/tex] -------------(iii)
In this case,
Ф = A
opposite = 24 ft (This is the opposite side to angle A)
adjacent = 36 ft (This is the adjacent side to angle A)
Substitute these values into equation (iii) as follows;
tan A = [tex]\frac{24}{36}[/tex]
tan A = [tex]\frac{2}{3}[/tex]
