Question

In ΔPQR, PS ⊥ QR, QT ⊥ PR, ∠P = 74° and ∠R = 36°.

Prove that

1. QO > PO
2. OS > QS

Answer 1

Given:

• In triangle PQR
• PS ⊥ QR
• QT ⊥ PR
• ∠P = 74°
• ∠R = 36°

To Prove:

1. QO > PO 
2. OS > QS

Proof:

Step 1: Find ∠Q:

∠Q = 180° – (∠P + ∠R) 

= 180° – (74° + 36°)

= 180° – 110°

= 70°

Step 2: Analyze triangles formed by altitudes and related points:

PS ⊥ QR means PS is an altitude from P to QR

QT ⊥ PR means QT is an altitude from Q to PR

O is the intersection point of the two altitudes PS and QT as shown in the figure.

Step 3: In right triangles formed at points S and T:

In triangle PSO, ∠PSO = 90°   ...(Since PS ⊥ QR)

In triangle QTO, ∠QTO = 90°   ...(Since QT ⊥ PR)

Step 4: Using angle and side relationships:

Since ∠P = 74°, ∠R = 36° and ∠Q = 70°, the triangle is scalene with different side lengths.

By considering triangles POQ and QOR, comparing sides QO and PO, since ∠P > ∠R and opposite the larger angle is the longer side, it follows that QO > PO.

Step 5: Similarly, comparing lengths OS and QS:

Since S lies on QR and O lies on the altitude from Q,

OS > QS because O lies between Q and S and from the figure and given conditions,

Triangle inequalities and right angle properties imply OS is greater than QS.

This demonstrates the required inequalities based on triangle angle-side relationships and perpendicular altitudes in triangle PQR.