# help mi wif tis question!!

• Jun 18th 2006, 08:24 PM
xiaoz
help mi wif tis question!!
help mi wif tis question below~
• Jun 19th 2006, 02:05 AM
ticbol
Quote:

Originally Posted by xiaoz
help mi wif tis question below~

1) First question.

1.i) Find QR.
Angle QPR = 90 -75 = 15 degrees.
By Law of Cosines,
(QR)^2 = (72)^2 +(115)^2 -2(72)(115)cos(15deg)
(QR)^2 = 2413.2683

1.ii) Find (90 +angle PRQ) degrees. ---------bearing is measured clockwise.
By Law of Sines,
sin(angle PRQ) /(72) = sin(15deg) /(49.125)
sin(angle PRQ) = [sin(15deg) /(49.125)]*(72) = 0.3793378
angle PRQ = arcsin(0.3793378) = 22.29 degrees
Therefore, bearing of Q from R is (90 +22.29) = 112.29 degrees -----answer.

----------------------

2.) Second Question.

2.i) Area of triangle PQR in terms of b, h and x.
In triangle PQR, the height is 20% higher than h, so, height of this new triangle is (1.20)h = 1.2h
Hence, area of triangle PQR = (1/2)(base)(height) = (1/2)(b+x)(1.2h)

2.ii) Find x, in terms of b, if area of triangle PQR is twice that of triangle ABC.

Area of triangle ABC = (1/2)(b)(h)
So,
(0.6)(b+x)(h) = 2[(1/2)(b)(h)]
(0.6)(b+x)(h) = b(h)
The h cancels out,
(0.6)(b+x) = b
(0.6)b +(0.6)x = b
(0.6)x = b -(0.6)b = (0.4)b
x = (0.4)b / (0.6)
• Jun 19th 2006, 03:01 AM
xiaoz
thxz.. but the ans for 1ii) is 292.3 degree.. ur ans is not the same ? why ?
• Jun 19th 2006, 04:27 AM
ticbol
Quote:

Originally Posted by xiaoz
thxz.. but the ans for 1ii) is 292.3 degree.. ur ans is not the same ? why ?

Huh?
I see.
I based my bearing from the South, that is why only 90 degrees was added to 22.29 degrees.
So the bearing really starts from the North, then go clockwise. So, North, +90deg first quadrant, +90deg 4th quadrant, +90deg 3rd quadrant, +22.29deg, equals (90+90+90+22.29) =?
• Jun 20th 2006, 12:25 AM
xiaoz
o... kkz hahax