Question

In ∆OPQ, right-angled at P, OP = 7 cm and OQ – PQ = 1 cm, then the values of sin Q.​

Answer 1

Given data : In ∆ OPQ, right-angled at P, OP = 7 cm and OQ – PQ = 1 cm. To find : The value of sin Q.

Solution : A/C to given data;

In right angle traingle, ∆ OPQ

• OP = 7 cm

➜ OQ - PQ = 1 cm

➜ OQ = 1 + PQ ----{1}

Here, a/c to given, OQ is hypotenuse

Now, we use Pythagoras theorem:

➜ (hypo)² = (first side)² + (second side)²

➜ (OQ)² = (OP)² + (PQ)²

➜ (1 + PQ)² = 7² + (PQ)²

➜ 1 + PQ² + 2PQ = 49 + PQ²

➜ PQ² - PQ² + 2PQ + 1 - 49 = 0

➜ 2PQ - 48 = 0

➜ 2 PQ = 48

➜ PQ = 48/2

➜ PQ = 24 cm

Put value of PQ in eq. {1}

➜ OQ = 1 + PQ

➜ OQ = 1 + 24

➜ OQ = 25 cm

Now, to find out, sin Q = ?

Here we know that; sin θ = opposite side/hypo

{A/C to figure;}

➜ sin Q = OP/OQ

➜ sin Q = 7/25

Answer : Hence, sin Q = 7/25.

Answer 2

Answer:

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Step-by-step explanation:

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