help mi wif tis question below~

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- Jun 18th 2006, 09:24 PMxiaozhelp mi wif tis question!!
help mi wif tis question below~

- Jun 19th 2006, 03:05 AMticbolQuote:

Originally Posted by**xiaoz**

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

QR = 49.125 meters ----------answer.

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)

= (0.6)(b+x)(h) ------------answer.

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)

x = (2/3)b -----------answer. - Jun 19th 2006, 04:01 AMxiaoz
thxz.. but the ans for 1ii) is 292.3 degree.. ur ans is not the same ? why ?

- Jun 19th 2006, 05:27 AMticbolQuote:

Originally Posted by**xiaoz**

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, 01:25 AMxiaoz
o... kkz hahax