# Thread: Basic Trigonometry

1. ## Basic Trigonometry

The triangle ABC has sides AB [] AC [] and BC [], as well as cosBaC []

a] Write down and simplify a quadratic equation satisfied by and hence evaluate

b] Express sinBaC in the form

c] Express sinAcB in the form

I have no idea how to solve these, it's the most difficult question I have out of the exam papers and even though I (sort of) understand trigonometry at the moment this is just impossible to me.

For a] I have a value of 8, and this I know to be right

For b] I have $\displaystyle \frac{sin73.4}{9} = \frac{sinB}{8}$ but I don't know how you can express it in the form asked for.

Please could someone help?

2. Originally Posted by db5vry
The triangle ABC has sides AB [] AC [] and BC [], as well as cosBaC []

a] Write down and simplify a quadratic equation satisfied by and hence evaluate

b] Express sinBaC in the form

c] Express sinAcB in the form

I have no idea how to solve these, it's the most difficult question I have out of the exam papers and even though I (sort of) understand trigonometry at the moment this is just impossible to me.

For a] I have a value of 8, and this I know to be right

For b] I have $\displaystyle \frac{sin73.4}{9} = \frac{sinB}{8}$ but I don't know how you can express it in the form asked for.

Please could someone help?

b) you can use the formula: sin^2 A + cos^2 A =1

c) in b) you will obtain sin BaC= (square of 45)/7. you know 7cm is one side of the triangle so you know that (square of 45) cm is the height. so sin AcB=(square of 45)/9 [because 9 is the other side

3. Hello, db5vry!

The triangle ABC has sides: $\displaystyle AB = 7,\;BC = 9,\;AC = x$ and $\displaystyle \cos A = \tfrac{2}{7}$
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B
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7  *     *   9
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A *  *  *  *  *  *  * C
x
(a) Write down and simplify a quadratic equation satisfied by $\displaystyle x$ and evaluate $\displaystyle x$

Law of Cosines: .$\displaystyle a^2 \:=\:b^2 + c^2 - 2bc\cos A$

We have: .$\displaystyle 9^2 \:=\:x^2 + 7^2-2(x)(7)\cos A \quad\Rightarrow\quad 81 \:=\:x^2 + 49 - 14x\left(\tfrac{2}{7}\right)$

. . which simplifies to: .$\displaystyle \boxed{x^2 - 4x - 32 \:=\:0}\quad\Rightarrow\quad (x-8)(x+4) \:=\:0$

Therefore: .$\displaystyle x \,=\,8\quad {\color{red}\rlap{///////}}x \,=\,-4$

(b) Express $\displaystyle \sin B$ in the form $\displaystyle \frac{\sqrt{m}}{n}$

Law of Sines: .$\displaystyle \frac{\sin B}{b} \:=\:\frac{\sin A}{a} \quad\Rightarrow\quad \sin B \:=\:\frac{b\sin A}{a}\;\;{\color{blue}[1]}$

We have: .$\displaystyle b = x = 8,\:a = 9$
And: .$\displaystyle \sin^2\!A \:=\:1-\cos^2\!A \:=\:1-\left(\tfrac{2}{7}\right)^2 \:=\:1 - \frac{45}{49} \quad\Rightarrow\quad \sin A \:=\:\frac{\sqrt{45}}{7}$

Substitute into [1]: .$\displaystyle \sin B \:=\:\frac{8\left(\frac{\sqrt{45}}{7}\right)}{9} \:=\:\frac{8\sqrt{45}}{63} \:=\:\frac{\sqrt{2880}}{63}$

(c] Express $\displaystyle \sin C$ in the form $\displaystyle \frac{\sqrt{P}}{3}$

Law of Sines: .$\displaystyle \frac{\sin C}{c} \:=\:\frac{\sin B}{b} \quad\Rightarrow\quad \sin C \:=\:\frac{c\sin B}{b}\;\;{\color{blue}[2]}$

We have: .$\displaystyle c = 7,\;b = x = 8,\;\sin B = \frac{8\sqrt{5}}{21}$

Substitute into [2]: .$\displaystyle \sin C \:=\:\frac{7\left(\frac{8\sqrt{5}}{21}\right)}{8} \:=\;\frac{\sqrt{5}}{3}$