Step 1
Draw a line segment AB of 5 cm. Taking A and B as centre, draw arcs of 6 cm and 5 cm radius respectively.
Let these arcs intersect each other at point C. ΔABC is the required triangle having length of sides as 5 cm, 6 cm, and 7 cm respectively.
Step 2
Draw a ray AX making acute angle with line AB on the opposite side of vertex C.
Step 3
Locate 7 points, A₁, A₂, A₃, A₄, A₅, A₆, A₇, (as 7 is greater between 5 and 7), on line AX such that AA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅ = A₅A₆ = A₆A₇.
Step 4
Join BA₅ and draw a line through A₇ parallel to BA₅ to intersect extended line segment AB at point B’.
Step 5
Draw a line through B’ parallel to BC intersecting the extended line segment AC at C’ ΔAB’C’ is the required triangle.
Justification
The construction can be justified by proving that
AB’ = 7/5 AB, B’C’ = 7/5 BC, AC’ = 7/5 AC
In ΔABC and Δ AB’C’,
∠ABC = ∠AB’C’ (Corresponding angles)
∠BAC = ∠B’AC’ (common)
∴ ΔABC ∼ ΔAB’C’ (AA similarity Criterion)
⇒ AB’/AB = B’C’/BC = AC’/AC …(1)
In ΔAA₅B/AA₇B’,
∠A₅AB = ∠A₇AB’ (common)
∠AA₅B = ∠AA₇B'(Corresponding angles)
∴ ΔAA₅B ∼ ΔAA₇B’ (AA similarity criterion)
⇒ AB’/AB = AA₅/AA₇ ⇒ AB’/AB = 5/7 …(2)
On comparing equations (1) and (2), we obtain.
AB’/AB = B’C’/BC = AC’/AC = 7/5
⇒ AB’ = 7/5 AB, B’C’ = 7/5 BC, AC’ = 7/5 AC
This justifies the construction.

Given: TQ and TP are two tangents of the circle.
We know that the radius is perpendicular to tangent. Therefore, OP ⊥TP and OQ TQ.
⇒ ∠OPT = 90° and ∠OQT = 90°
In quadrilaterral POQT, OQT + OQT + PTQ = 360°
⇒ 90° + 110° + 90° + ∠PTQ = 360°
⇒ ∠PTQ = 360° – 290° = 70
Hence, the option (B) is correct.

Let AB is diameter, PQ and RS are tangents drawn at ends of diameter.
We Know that the radius is perpendicular to tangent. therefore, OP ⊥PQ.
∠OAR = 90° and OAS = 90°
∠OBP = 90° and ∠OBQ = 90°
From the above, we have
∠OAR = ∠OBQ [Alternate angles]
∠OAS = ∠OBP [Alternate angles]
Since, alternate angles are equal. Hence, PQ is parallel to PS.