Question

In a triangle pqr p =120 ps is the perpendicular on qr pq+qs= sr find the angle of q

Answer 1

The angle of Q will be $\bold {40^o}$

Step-by-step explanation:

Let PQ = x

and QS = y

Now, we can take a point T on SR where ST = QS = y,

1. Now in triangle PQT (as mentioned Below)

$PS \perp QT \ \textrm{and} \ QS = ST$

so that triangle PQT is an isosceles triangle.

2. Now PQ = PT = x = TR,

Then PTR is also isosceles triangle.

Let angle $\angle R = \angle TPR = \theta \\\textrm{then,}\\ \angle PTQ = 2\theta \textrm{exterior angle of \Delta PTR} \\\textrm{Now},\\\angle PTQ = \angle PQT  = 2\theta$

$\therefore \angle QPT = 180\degree - (2\theta +2\theta )\\=180\degree - (4\theta )\\NOW,\\\angle QPR = \angle QPT + \angle TPR\\120^o = 180^o - 4\theta + \theta \\\theta = 20^o$

Now,

$\angle PQT = 2\theta = 2\times 20^o = 40^o\\\bold {\angle Q = 40^o}$