Anyone knows how to solve Q2?
1. In ABC, A=34 16', a=9 and c=12. Find the two possible values of b.
2. A triangle ABC lies on a horozontal plane with X, Y and Z vertically above A,B and C respectively. M is the mid-point of YZ, AB=5cm, BC=6cm, XA=YB=ZC=8cm, and ABC=90 . Find the angle between
a) the line Am and the plane ABC
b) the line AB and the line XC
1. 15.9, 3.97
a) 53 55'
b) 63 26'
Ok then, let's go through these step by step.
Q1... you are not told this is right-angled, so let's assume it's not.
You use the Sine Rule when you are given a triangle side length and it's opposite angle (or vice versa).
Side 'a' is opposite angle 'A'.
Side 'b' is opposite angle 'B'.
Side 'c' is opposite angle "C".
That's the convention, little letters for side lengths, capitals for angles.
You are given 'a' and 'A", so we can use the Sine Rule
if we are given a 3rd measurement, which we are.
We first find SinC, then we get C from that.
Then we can find B and finally b.
Now, since all 3 angles in a triangle add up to
to find the possible values of b.
You need to know that a triangle can have an angle and
Try that first.
Here is a diagram of Q2.
Imagine that ABC is lying flat on the floor and XYZ is identical to it,
but 8 units directly above it.
If you find the midpoint of YZ and draw a vertical line down onto BC,
you have two right angled triangles Asm and ABs.
ABs is a right-angled triangle.
Can you use Pythagoras' theorem to find |As| ?
Then you can calculate the angle sAm using inverse tangent.