For question 4, [tex]sinA =\frac{a}{c}, cosA= \frac{b}{c}, tan B = \frac{b}{a}, sin J = \frac{j}{l}, cosK = \frac{j}{l}, tanK = \frac{k}{j}.[/tex]
Question 5. Option a and question 6. Option j
Step-by-step explanation:
Step 1:
The three basic formula needed to solve these questions are:
[tex]sin\theta = \frac{oppositeside}{hypotenuse} , cos\theta = \frac{adjacentside}{hypotenuse}, tan\theta= \frac{opposite side}{adjacent side}.[/tex]
Step 2:
Using the above formula, we solve the following values
[tex]sinA = \frac{oppositeside}{hypotenuse}[/tex] [tex]=\frac{a}{c}.[/tex]
[tex]cosA = \frac{adjacentside}{hypotenuse}[/tex] [tex]= \frac{b}{c}.[/tex]
[tex]tanB= \frac{opposite side}{adjacent side}[/tex][tex]= \frac{b}{a}.[/tex]
[tex]sinJ = \frac{oppositeside}{hypotenuse}[/tex] [tex]= \frac{j}{l}.[/tex]
[tex]cosK = \frac{adjacentside}{hypotenuse}[/tex][tex]= \frac{j}{l}.[/tex]
[tex]tanK= \frac{opposite side}{adjacent side}[/tex][tex]= \frac{k}{j}.[/tex]
Step 3:
For question 5, The triangle's angle = 23°, opposite side = BC inches and hypotenuse = 4 inches.
[tex]sin\theta= \frac{opposite side}{hypotenuse}. sin 23^{\circ}= \frac{BC}{4}, sin23^{\circ} = 0.3907,BC = (0.3907)(4) = 1.5628.[/tex]
SO BC is 1.5628 inches, rounding this off to the nearest tenth, we get BC = 1.6 inches which is option a.
Step 4:
For question 6, The triangle's angle = 50°, opposite side = QR m the adjacent side = 8.1 m.
[tex]tan\theta= \frac{opposite side}{adjacentside}. tan 50^{\circ}=\frac{QR}{8.1}, tan50^{\circ} = 1.1917,QR = (1.1917)(8.1) = 9.65277[/tex]
SO QR is 9.65277 meters, rounding this off to the nearest tenth, we get QR = 9.7 inches which is option j.
