Sine, cosine rule, need help with Q. (C2)

This has been bugging me for quite a bit..

In [triangle]ABC, AB = 10cm, BC = a[square root of 3]cm, AC = 5[square root of 13] and ABC = 150 degrees. (B = 150)

Calculate:

a) the value of a.

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I have tried putting all the info given into the sine, cosine rules and trying to end up with a quadratic or something to end up with a, but to no avail.

Anybody have an idea on how to do this?

Reply 1

insparato

15

Okay ive done this one before.

Using the cosine rule

Cos A = b^2 + c^2 -a^2/ 2bc

cos 150 = 10^2 + (aroot3)^2 - (5root13)^2/2(10 x aroot3)

cos 150 = 100 + 3a^2 - 325/20aroot3

cos 150 = 3a^2 - 225/20aroot3

cos 150 = -root3/s

-root3/2 = 3a^2 - 225/20aroot3

20aroot3(-root3)/2 = 3a^2 -225

-60a/2 = 3a^2 - 225

-30a = 3a^2 - 225

3a^2 + 30a - 225 = 0

divide both sides by 3 to make life easier factorising.

a^2 + 10a - 75 = 0

(a - 5)(a + 15)

a = 5 a = -15

so a must = 5

Reply 2

Bengaltiger

Use the cosine rule:

(5\sqrt13)^2=10^2 + (a\sqrt3)^2 - 2(10)(a\sqrt3)cos 150

Then simplify.

Reply 3

dynamite_man

OP

Thanks both. [+Rep x2] (Will rep one tommorow).

Reply 4

Agent Solid

4

Original post by insparato

Okay ive done this one before.

Using the cosine rule

Cos A = b^2 + c^2 -a^2/ 2bc

cos 150 = 10^2 + (aroot3)^2 - (5root13)^2/2(10 x aroot3)

cos 150 = 100 + 3a^2 - 325/20aroot3

cos 150 = 3a^2 - 225/20aroot3

cos 150 = -root3/s

-root3/2 = 3a^2 - 225/20aroot3

20aroot3(-root3)/2 = 3a^2 -225

-60a/2 = 3a^2 - 225

-30a = 3a^2 - 225

3a^2 + 30a - 225 = 0

divide both sides by 3 to make life easier factorising.

a^2 + 10a - 75 = 0

(a - 5)(a + 15)

a = 5 a = -15

so a must = 5

WHY is it 3a^2 shouldnt it be just be 3a?

Reply 5

insparato

15

(a\sqrt3)^2 = 3a^2