Answer:
1) BC = 6
2) B = 33.7° (nearest tenth)
3) [tex]\sf AB=2\sqrt{13}[/tex]
Step-by-step explanation:
Trigonometric ratios
[tex]\sf \sin(\theta)=\dfrac{O}{H}\quad\cos(\theta)=\dfrac{A}{H}\quad\tan(\theta)=\dfrac{O}{A}[/tex]
where:
- [tex]\theta[/tex] is the angle
- O is the side opposite the angle
- A is the side adjacent the angle
- H is the hypotenuse (the side opposite the right angle)
Question 1
Given:
- Angle = A
- Side opposite angle = BC
- Side adjacent angle = AC = 4
Substituting the values into the tan ratio:
[tex]\implies \sf \tan A=\dfrac{BC}{AC}=\dfrac{BC}{4}[/tex]
Given:
[tex]\sf \tan A=\dfrac{3}{2}[/tex]
Therefore:
[tex]\implies \tan \sf A = \tan \sf A[/tex]
[tex]\implies \sf \dfrac{BC}{4}=\dfrac{3}{2}[/tex]
[tex]\implies \sf BC=\dfrac{4 \cdot 3}{2}=6[/tex]
Question 2
Given:
- Angle = B
- Side opposite angle = AC = 4
- Side adjacent angle = BC = 6
Substituting the values into the tan ratio and solving for B:
[tex]\implies \sf \tan B=\dfrac{AC}{BC}[/tex]
[tex]\implies \sf \tan B=\dfrac{4}{6}[/tex]
[tex]\implies \sf B=\tan ^{-1}\left(\dfrac{4}{6}\right)[/tex]
[tex]\implies \sf B=33.69006753...^{\circ}[/tex]
[tex]\implies \sf B=33.7^{\circ}\:(nearest\:tenth)[/tex]
Question 3
Pythagoras’ Theorem
[tex]a^2+b^2=c^2[/tex]
(where a and b are the legs, and c is the hypotenuse, of a right triangle)
Given:
- a = AC = 4
- b = BC = 6
- c = AB
Substituting the values into the formula and solving for AB:
[tex]\implies \sf AC^2+BC^2=AB^2[/tex]
[tex]\implies \sf 4^2+6^2=AB^2[/tex]
[tex]\implies \sf 16+36=AB^2[/tex]
[tex]\implies \sf AB^2=52[/tex]
[tex]\implies \sf AB=\sqrt{52}[/tex]
[tex]\implies \sf AB=2\sqrt{13}[/tex]
