Please help me. C4 trignometerey question part 4
I have a question as follows. The diagram shows 3 points L(-2,1), M(0,2) and N(3,-2) joined to form a triangle. The angles a and b and the point P are shown in the diagram.
angle a is left of x=0 and angle b is right of x=0 - making up compound angle of a + b. ...
P is at (0, 1)
Question (i) asks Show that sin a =2/root 5 and write down the value of cos a.
The next part, (ii) asks Find the values of sin b and cos b.
iii) show that sin LMN = 11/5 root 5
iv) show that tan LNM = 11/27.
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C4 trig please help!!! watch
- Thread Starter
- 28-01-2017 18:30
- 28-01-2017 20:20
i) using trig and pythagoras it is easy to find out that LM = root(5), LP = 2, and MP = 1, therefore the trig values are simple for triangle LPM
ii) Say the line MN crosses the x axis at K. Use triangle MOK to work out trig values for B. You know MO = 2, and once you've worked out K then OK = K. Then just work out trig values.
iii) use sin(a+b) = sin(a)cos(b) + sin(b)cos(a) and your previous answers.
iv) sin = opposite/hypoternuse, so construct a triangle with the side lengths of the answer to (iii) and use pythagoras.
Let me know if you need anymore help
- 28-01-2017 20:24
jjust clarification for iv, tan = opposite/adjacent and sin = opposite/hypoternuse
use the answer they give you to (iii) to work out the value of the adjacent side using pythagoras, then you'll be able to give an answer for tan.
- 28-01-2017 20:39
For part (iii) you could also wheel out the cosine rule as you already have (or can easily find) the lengths of all three sides of the triangle LNM. This will give you cos(LMN). Sin (LMN) will follow via a quick sketch of a triangle with adjacent and hypotenuse to suit your cos(LMN). Pythagoras will give you the missing side (opposite) and you can then find sin(LMN).