Hi:) how to do this?

b) 15.2km

c) Bearing= 263.8° (1 d.p.),

AD= 25.5 km (3 s.f.)

Step-by-step explanation:

Please see attached picture for full solution.

In picture 3, you could also use trigonometry to find AW, just like how we did to find W.

Picture 4:

WD= 16 -15.250

WD= 0.75

Picture 5:

Bearing of D from A

= reflex ∠N₂AD

= 295° - ∠BAD

∠BAD

= ∠BAW -∠WAD

= 33° -1.829°

= 31.171°

Thus, bearing of D from A

= 295° - 31.171°

= 263.8° (1 d.p.)

c) Distance of D from A= length AD

Look at triangle BAD:

∠DBA= 65°

∠ADB= 180° -65° -31.171° (∠ sum of △ADB)

∠ADB= 83.829°

Using sine rule,

Feel free to ask if you have any queries :)

if gloria has "carry over" minutes, (that is, any minutes paid for but not used in a month can be applied to her payment for minutes in the next month,) her total usage for the three months is: total minutes = 320 + 243 + 489 = 1052 minutes number of 200 minute blocks = 1052/200 = 5 r52 (that is 5 with a remainder of 52) rounding up the number of 200 minute blocks gives her usage for the three months as 6 blocks. the cost for six blocks is 6 x $25 = $150 she also has 200 - 52 or 148 carry over minutes if gloria pays each month and unused minutes are lost, applying the rounding up procedure to each month gives: first month usage of 320 minutes rounds up to 2 blocks of 200 minutes for a payment of 2 x $25 = $50 second month usage of 243 minutes rounds up to 2 blocks of 200 minutes for a payment of 2 x $25 = $50 third month usage of 489 minutes rounds up to 3 blocks of 200 minutes for a payment of 3 x $25 = $75 total payments for the three months are $50 + $50 + $75 = $175 get the plan with carry over minutes!

what is the your question?