Question

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Answer 1

Given that PQ + QR = 30cm.

According to Pythagoras Theorem,

PQ² + QR² = PR²

$\longrightarrow$ 30² + 30² = PR²

$\longrightarrow$ PR² = 2(30)²

$\longrightarrow$ PR = √2(30)²

$\longrightarrow$ PR = 30√2 cm.

Further,

• sin∅ = Opposite side/Hypotenuse
• sec∅ = Hypotenuse/Adjacent Side

Now,

sin²R = (PQ/PR)²

$\longrightarrow$ sin²R = (30/30√2)²

$\longrightarrow$ sin²R = (1/√2)²

$\longrightarrow$ sin²R = 1/2

Also,

sec²R = (PR/QR)²

$\longrightarrow$ sec²R = (30√2/30)²

$\longrightarrow$ sec²R = (√2)²

$\longrightarrow$ sec²R = 2

Therefore,

sin²R + sec²R

$\longrightarrow$ 1/2 + 2

$\longrightarrow$ (1 + 4)/2

$\longrightarrow$ 5/2

Hence, Proved.

Answer 2

Given that PQ + QR = 30cm.

According to Pythagoras Theorem,

PQ² + QR² = PR²

$\sf \bf⟶ 30² + 30² = PR²$

$\sf \bf⟶ PR² = 2(30)²$

$\sf \bf⟶ PR = √2(30)²$

$\bf \sf⟶ PR = 30√2 cm.$

Further,

• sin∅ = Opposite side/Hypotenuse
• sec∅ = Hypotenuse/Adjacent Side

Now,

sin²R = (PQ/PR)²

⟶ sin²R = (30/30√2)²

⟶ sin²R = (1/√2)²

⟶ sin²R = 1/2

Also,

sec²R = (PR/QR)²

⟶ sec²R = (30√2/30)²

⟶ sec²R = (√2)²

⟶ sec²R = 2

Therefore,

sin²R + sec²R

⟶ 1/2 + 2

⟶ (1 + 4)/2

⟶ 5/2

Hence, Proved.