Question

Can anyone do 4,5 and 6 for me plz

Answer 1

For question 4,
`sinA =(a)/(c), cosA= (b)/(c), tan B = (b)/(a), sin J = (j)/(l), cosK = (j)/(l), tanK = (k)/(j).`

Question 5. Option a and question 6. Option j

Explanation:

Step 1:

The three basic formula needed to solve these questions are:

`sin\theta = (oppositeside)/(hypotenuse) , cos\theta = (adjacentside)/(hypotenuse), tan\theta= (opposite side)/(adjacent side).`

Step 2:

Using the above formula, we solve the following values

`sinA = (oppositeside)/(hypotenuse)`
`=(a)/(c).`

`cosA = (adjacentside)/(hypotenuse)`
`= (b)/(c).`

`tanB= (opposite side)/(adjacent side)`
`= (b)/(a).`

`sinJ = (oppositeside)/(hypotenuse)`
`= (j)/(l).`

`cosK = (adjacentside)/(hypotenuse)`
`= (j)/(l).`

`tanK= (opposite side)/(adjacent side)`
`= (k)/(j).`

Step 3:

For question 5, The triangle's angle = 23°, opposite side = BC inches and hypotenuse = 4 inches.

`sin\theta= (opposite side)/(hypotenuse). sin 23^(\circ)= (BC)/(4), sin23^(\circ) = 0.3907,BC = (0.3907)(4) = 1.5628.`

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.

`tan\theta= (opposite side)/(adjacentside). tan 50^(\circ)=(QR)/(8.1), tan50^(\circ) = 1.1917,QR = (1.1917)(8.1) = 9.65277`

SO QR is 9.65277 meters, rounding this off to the nearest tenth, we get QR = 9.7 inches which is option j.