Question

Please yaar please help me ​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

PS bisects ∠QPR

⟹ ∠QPS = ∠SPR = 'x' say

Let assume that ∠ TPS = 'y'.

So, ∠QPT = ∠QPS - ∠TPS = x - y

Also, ∠RPT = ∠SPR + ∠TPS = x + y

Now, In ∆ PTQ

It is given that PT ⊥ QR

⟹ ∠PTR = 90°

We know, sum of all interior angles of a triangle is supplementary.

So, ∠PQT + ∠PTQ + ∠TPQ = 180°

⟹ ∠Q + x - y + 90° = 180° ---------(1)

Now, In ∆ PTR

It is given that PT ⊥ QR

⟹ ∠PTR = 90°

We know, sum of all interior angles of a triangle is supplementary.

So, ∠TRQ + ∠PTR + ∠TPR = 180°

⟹ ∠R + x + y + 90° = 180° ---------(2)

On equating equation (1) and equation (2), we get

$\rm :\longmapsto\:\angle Q + x - y + 90 \degree \: = \: \angle R \: + x + y + 90 \degree$

$\rm :\longmapsto\:\angle Q - y = \: \angle R + y$

$\rm :\longmapsto\:\angle Q - \angle R = 2y$

$\rm :\longmapsto\:y \: = \: \dfrac{1}{2} (\angle Q - \angle R)$

$\bf\implies \:\:\angle TPS \: = \: \dfrac{1}{2} (\angle Q - \angle R)$

Hence, Proved

Properties of a triangle

Angle Sum Property of triangle :- The sum of all interior angles of a triangle is supplementary. 

The sum of two sides of a triangle is always greater than the third side.

The side opposite to the largest angle of a triangle is the largest side.

The angle opposite to greatest side is always larger.

Exterior angle Property of the triangle :- Exterior angle of a triangle is equal to the sum of its interior opposite angles.

Based on the angle measurement, there are three types of triangles:

Acute Angled Triangle : A triangle having all three angles less than 90° is an acute angle triangle.

Right-Angled Triangle : A triangle that has one angle 90° is a right-angle triangle.

Obtuse Angled Triangle : A triangle having one angle more than 90° is an obtuse angle triangle.