# Addition Formula Question

• Jan 13th 2011, 09:21 AM
cromlix
Hi there,
I have a problem involving a right angled triangle ABC. It is right angled at C.
The hypoteneuse has a value of 13 units.
A line stretches across the triangle from B to side AC at D. Side AC is divided into AD and DC. DC has the value 4 units.
Angle ABD equals x degrees and angle DBC equals 30 degrees.
I have to find the exact value of sin x.

I have worked out that side BC has a value of 6.928 units. Hypoteneuse BD has a value of 7.9 units and side AD equals 7 units.
Can anyone give me help with this problem please?

Thank you,

Cromlix
• Jan 13th 2011, 10:18 AM
Quote:

Originally Posted by cromlix
Hi there,
I have a problem involving a right angled triangle ABC. It is right angled at C.
The hypoteneuse has a value of 13 units.
A line stretches across the triangle from B to side AC at D. Side AC is divided into AD and DC. DC has the value 4 units.
Angle ABD equals x degrees and angle DBC equals 30 degrees.
I have to find the exact value of sin x.

I have worked out that side BC has a value of 6.928 units. Hypoteneuse BD has a value of 7.9 units and side AD equals 7 units.
Can anyone give me help with this problem please?

Thank you,

Cromlix

In triangle BCD

$\displaystyle tan30^o=\frac{4}{|BC|}\Rightarrow\ |BC|=4\sqrt{3}$

By Pythagoras' Theorem on triangle ABC

$\displaystyle \left[4\sqrt{3}\right]^2+|AC|^2=13^2=169\Rightarrow\ |AC|=11\Rightarrow\ |AD|=7$

Now you can use the Sine Law on triangle ABD

$\displaystyle \frac{Sin\;x}{7}=\frac{Sin120^o}{13}$

Angle ADB is $\displaystyle 120^o$ because angle DCB is $\displaystyle 90^o-30^o$
• Jan 13th 2011, 10:21 AM
emakarov
Too late...

I think the easiest thing is to use the law of sines for the triangle ABD. The angle ADB is $\displaystyle 90+30=120^\circ$. Therefore, $\displaystyle AD/(\sin x)=AB/(\sin 120)$.

A longer way is to say $\displaystyle \cos(30+x)=BC/AB$ and rewrite $\displaystyle \cos(30+x)$ as $\displaystyle \cos30\cos x-\sin 30\sin x$. Since $\displaystyle x<90$, we have $\displaystyle \cos x>0$ and so $\displaystyle \cos x=\sqrt{1-\sin^x}$. This way we get a quadratic equation on $\displaystyle \sin x$.
• Jan 13th 2011, 10:28 AM
Opalg
Quote:

Originally Posted by cromlix
I have a problem involving a right angled triangle ABC. It is right angled at C.
The hypotenuse has a value of 13 units.
A line stretches across the triangle from B to side AC at D. Side AC is divided into AD and DC. DC has the value 4 units.
Angle ABD equals x degrees and angle DBC equals 30 degrees.
I have to find the exact value of sin x.

I have worked out that side BC has a value of 6.928 units. Hypotenuse BD has a value of 7.9 units and side AD equals 7 units.

The values you have calculated for BC and BD are approximately correct. But the question is asking for exact values. So instead of using a calculator, you should have used the fact that $\displaystyle \tan 30^\circ = 1/\sqrt3$ to get the exact value BC = $\displaystyle 4\sqrt3$, as in Archie Meade's comment. Also, BD is exactly 8 (not 7.9). But you are correct to say that AD = 7, and this is exact (because AC = 11 by Pythagoras, and when you subtract 4 for DC you are left with exactly 7 for AD). Now you can complete the question as in Archie's comment.