Cosine Rule (1 Viewer)

[Mc_Meaney]

ABC is a triangle in which AB=3cm, BC=6cm and CA= 5cm.

(i) Use the cosine rule to find angle BAC to the nearest degree

(ii) Calcullate the area of triangle ABC to 2sig fig.

For (I) do use the formula 2bcCosA....i have no idea really lol help please someone

as for (ii) lol...nfi

 

[Riviet]

let A be the angle we're finding (^BAC).

cosA=(b²+c²-a²)/2bc, where b and c are the sides AB and AC, and a is BC. Use this formula to find an angle.

If asked to find a side, use

a²=b²+c²-2bc.cosA

For area of triangle, use

Area=1/2.bc.sinA, using the same angles and sides mentioned before.

 

[SoulSearcher]

Let angle BAC = x
cos x = (3² + 5² - 6²) / (2 * 5 * 3)
cos x = -2 / 30 = -1 / 15
therefore x = cos^-1 (-1/15)
= 93.8 degress = 94 degrees to the nearest degree.

For part 2, use the anlge you have just found, so
Area of ABC = 1/2 * 3 * 5 * sin 94
= 7.5 cm², note that I am using the exact answer for part (1) as the angle for part 2

 

[Mc_Meaney]

Thanx guys

 

[Mc_Meaney]

1 quick question though, Tanjent to a curve - you differentiate and plug in a point yes? But how do I find the equation....its arghhhh

so for y= x^2+5/x-2 at (1,4)

 

[SoulSearcher]

Yes.

y = x² + 5/(x-2)
y' = 2x - 5/(x-2)²
when x = 1, y' = 2 - 5/(1-2)²
= 2 - 5/(-1)²
= 2 - 5
= -3
therefore equation of tangent is
y - 4 = -3(x - 1)
y - 4 = -3x + 3
3x + y - 7 = 0

 

[Riviet]

In general, to find the equation of the tangent/normal, differentiate first, sub. in your x value to find the gradient at that point, then finally use y-y₁=m(x-x₁) to find the equation of it, where (x₁,y₁) is your point and m is the gradient.

 

[Mc_Meaney]

Thanx again

 

[Mc_Meaney]

Random Q= congruence = SSS SAS AAS RHS or is that similair trianglies/....

 

[pLuvia]

Similar triangles is AAA
Congruence is SAS, SSS, AAS, RHS

 

[Mc_Meaney]

thanx g

 

[Mc_Meaney]

(i) THere s one turnin point for f(x) = x - 3Inx. Find the coordinates and is it Max or Min.

I think ive done it right so far -

F(x) = x-3Inx
ddx = -3/x

But I have no idea if its right or the next step, because equating to 0 is impossible isnt it?

 

[SoulSearcher]

you forgot the derivative of x, which is 1, so that
d/dx(x- 3lnx) = 1 - 3/x
therefore when d/dx(x - 3ln) = 0
0 = 1 - 3/x
3/x = 1
x = 3, then you find the y coordinate.
To find the nature of the turing point, find the second derivative, which is 3/x², sub in x = 3 and you can figure it out from there.

 

[Mc_Meaney]

That is a very good point!!! oops