$P$ is a point inside triangle $ABC$ such that $\angle ABP=20^{\circ} $ , $\angle PBC=10^{\circ}$, $∠ ACP = 20°$ and $∠ PCB = 30°$. Determine $∠CAP$ , in degree.
No figure was given
I used the sine Law but the could not solve equations I got, I also tried making a point $D$ on $BP$ such that $BD = CD$ to make us of the $10^{\circ}$, but I still cant solve it.
On
The angle measure can be found by solving the 3 separate equations obtained by using the sine rule on $\triangle$s ABP, APC & PBC .
Using sine rule on △ ABP , we get
$$\frac{AP}{BP}= \frac{\sin 20}{\sin (100-x)}$$
Using sine rule on △s APC & BPC , and eliminating PC from the equations , we get
$$\frac{AP}{BP}= \frac{\sin 10 * \sin 20 }{\sin x * \sin 30} $$
From the two equations , we get $$\frac{\sin x}{2}= \sin (100-x)*\sin 10$$
Rewriting $ \sin (100-x)$ as $\cos (10-x)$ , and using the cosine sum formula, we get
$$\cot x = \frac{1-2\sin^2(10)}{2\sin10*\cos10} $$
Since $1 - 2\sin^2(t) = \cos 2t $ and $ 2\sin t * \cos t = \sin 2t $, we get
$$\cot x = \cot 20 $$ which implies $x$ is $\boxed{20°}$
On
Rotate $P$ for $-60^{\circ}$ around $B$ to new point $D$. Then $D$ is a center of a circumcircle for triangle $BCP$ so $$\angle CDP = 2\angle CBP = 20^{\circ}$$
So $ABCD$ is cyclic since $$\angle CDB +\angle CAB = 180^{\circ}$$
$ABCD$ is also a trapez since $$\angle CBP = \angle ACB = 50^{\circ}$$
So $AB = DC = BP$ and thus $\angle BAP = \angle BPA = 80^{\circ}$ and there for $$\angle CAP = 20^{\circ}$$
Extend BP so that it intersects AC at D and extend CP so that it intersects AB at E. Now consider triangle ADP and triangle AEP these triangle are congruent so angle APD =70 and angle DEC=40 APD+DEC=APC=110 Hence angle CAP=30