ABCD is a parallelogram a point is taken inside it and vertices are joined to it angle are marked as given in diagram find unknown angle

Question

ABCD is a parallelogram a point is taken inside it and vertices are joined to it angle are marked as given in diagram find unknown angle

Answer 1

Let $\alpha$ be another unknown angle in the diagram. Then, all the angles of the triangle ADP are known in terms of $\alpha$. Apply the sine rule to the triangles DPA, DPC and CPB,

$$\frac{\sin60}{\sin\alpha}= \frac{AD}{DP},\>\>\>\>\> \frac{\sin\alpha}{\sin40}= \frac{DP}{CP},\>\>\>\>\> \frac{\sin\theta}{\sin(160-\theta)}= \frac{CP}{CB}$$

Recognize AD = CB, $\sin(160-\theta) = \sin(\theta+20)$ and multiply the three ratios above to get,

$$\sin60\sin\theta=\sin40\sin(\theta+20)$$

or

$$\cos(60-\theta)-\cos(60+\theta) = \cos(\theta-20)-\cos(60+\theta) $$

which reduces to

$$\cos(60-\theta) = \cos(\theta-20)$$

and yields the solution $\theta = 40^\circ$.

Answer 2

Let that point inside be $P$ and translate it by vector $\vec{AB}$ to new point $Q$.

Then ABQP and PQCD are paralelograms so $$\angle PQB = 20^{\circ} = \angle BCP$$ so quadrilateral $BQCP$ is cyclic and thus $$\theta =\angle PBC = \angle PQC = 40^{\circ} $$