snore
Kinky Kelly & Sexy Stud
I was trying to do this question, 25 (b), but i just couldn't get what I thought was the right answer. I ended up a angle of elevation from M to K of 72 but I don't think that's right. Please help me.
Help with a 2007 HSC Exam Question (1 Viewer)
- Thread starter snore
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*Baby-K*
Member
I have the whole paper with worked answers if you want it PM me your email address and I'll send it to you cause it's too big for an attachment on BOS
*Baby-K*
Member
Yep, sent them to you. Check your email
Try to work out as many angles as you can before getting started.
^JML = 75° because it's an alternate angle to the given angle of depression.
^MJL = 15° because it makes up the 90° angle with the angle of depression (or it makes up the 180° angle sum in ∆JML).
Now in ∆JMK, we have ^MJK = 15°, MK = 18 m and JK = 20 m.
So using the sine rule:
JK / sin ^JMK = MK / sin ^MJK
20 / sin ^JMK = 18 / sin 15°
Cross multiplying gives:
18 sin ^JMK = 20 sin 15°
sin ^JMK = 20 sin 15° / 18
sin ^JMK = 0.2875767168
^JMK = 17°
Now we know that ^JML = 75° and now ^JMK = 17°, so ^KML = 75 – 17 = 58°, which is the angle of depression required.
^JML = 75° because it's an alternate angle to the given angle of depression.
^MJL = 15° because it makes up the 90° angle with the angle of depression (or it makes up the 180° angle sum in ∆JML).
Now in ∆JMK, we have ^MJK = 15°, MK = 18 m and JK = 20 m.
So using the sine rule:
JK / sin ^JMK = MK / sin ^MJK
20 / sin ^JMK = 18 / sin 15°
Cross multiplying gives:
18 sin ^JMK = 20 sin 15°
sin ^JMK = 20 sin 15° / 18
sin ^JMK = 0.2875767168
^JMK = 17°
Now we know that ^JML = 75° and now ^JMK = 17°, so ^KML = 75 – 17 = 58°, which is the angle of depression required.
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