1. ## Triangle

Anyone knows how to solve Q2?

1. In $\triangle$ABC, $\angle$A=34 $\circ$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 $\angle$ABC=90 $\circ$. 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

2.
a) 53 $\circ$55'
b) 63 $\circ$26'

2. $Are\ you\ looking\ for\ the\ methods\ to\ solve\ these?$

3. Originally Posted by Archie Meade
$Are\ you\ looking\ for\ the\ methods\ to\ solve\ these?$
Yes. Please? I've been thinking for hours!

4. $How\ are\ you\ getting\ on\ at\ your\ classwork?$

$How\ familiar\ are\ you\ with\ using\ Sine,\ Cosine\ and\ Tangent?$

$The\ formulae\ for\ area\ of\ a\ triangle,\ the Sine Rule\ and\ Cosine\ Rule?$

$Your\ first\ question\ is\ testing\ whether\ or\ not\ you\ recognise\ it\ as\ an\ application\$

$of\ the\ Sine\ Rule.$

$The\ 2nd\ question\ is\ testing\ you\ on\ Pythagoras' theorem$

$followed\ by\ inverse\ tan.$

5. Originally Posted by Archie Meade
$How\ are\ you\ getting\ on\ at\ your\ classwork?$

$How\ familiar\ are\ you\ with\ using\ Sine,\ Cosine\ and\ Tangent?$

$The\ formulae\ for\ area\ of\ a\ triangle,\ the Sine Rule\ and\ Cosine\ Rule?$

$Your\ first\ question\ is\ testing\ whether\ or\ not\ you\ recognise\ it\ as\ an\ application\$

$of\ the\ Sine\ Rule.$

$The\ 2nd\ question\ is\ testing\ you\ on\ 'triangle\ area'=0.5abSinC,$

$followed\ by\ the\ Cosine\ Rule.$
To be honest... I'm not very good in math but I did try those questions... What makes me mad is the answer I got was different from the given answer... For example, the answer is 55.53 but I got 61.85 instead...

6. You should try to post your working; to make it easier to see where you went wrong.

7. 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.

$\frac{SinA}{a}=\frac{SinC}{c}$

We first find SinC, then we get C from that.
Then we can find B and finally b.

$\frac{Sin34.27^o}{9}=\frac{SinC}{12}$

$\frac{0.563}{9}(12)=SinC$

$SinC=0.75,\ so\ C=Sin^{-1}(0.75)=48.6^o\ or\ (180-48.6)^o=131.4^o$

Now, since all 3 angles in a triangle add up to $180^o,$

$B=180-(48.6+34.27)=97.13^o\ or\ B=180-(131.4+34.27)=14.33^o$

Finally, use $\frac{b}{SinB}=\frac{a}{SinA}\ or\ \frac{c}{SinC}$

to find the possible values of b.
You need to know that a triangle can have an angle $>90^o$ and $<180^o$

Try that first.

8. I got it. Thank you so much!

9. Originally Posted by Archie Meade
Try that first.
Can you show me the method for Q2 also?

10. 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.

11. The line AB is on the plane ABC,
so I'm assuming the second part of Q2 is asking for the angle the line XC
makes with the ABC plane.

Use Pythagoras' theorem to calculate the length of the line AC.
Then use inverse tangent to calculate the angle ACX.

12. Originally Posted by Archie Meade
Use Pythagoras' theorem to calculate the length of the line AC.
Then use inverse tangent to calculate the angle ACX.
Oh I see... Thanks a lot. You're my life savior~