1

Problem

In the ∆OPQ, right-angled at P, PQ = 5cm, and OQ - OP = 3cm. Determine the value of sin Q + cos Q.

2

Step

In a right-angled triangle ∆OPQ, we can use the Pythagorean theorem and trigonometric ratios to find sin Q + cos Q.

3

Step

The given values are

PQ = 5 cm, OQ - OP = 3cm

4

Step

First, let’s find the length of the sides OQ and OP using the Pythagorean theorem:

\$ OQ^2 = OP^2 + PQ^2 \$

\$ OQ^2 = OP^2 + 5^2 \$

5

Step

Now, since OQ − OP = 3, we can express OQ in terms of OP:

OQ = OP + 3

6

Step

Now, substitute this into the Pythagorean theorem:

\$ (OP + 3)^2 = OP^2 + 5^2 \$

7

Step

Expand and simplify the equation:

\$ OP^2 + 9 + 6OP = OP^2 + 25 \$

\$ 9 + 6OP = 25 \$

8

Step

After simplification

\$ 6OP = 16 \$

\$ OP = (16)/6 = 8/3 \$

9

Step

Now that we have the value of OP, we can find OQ:

\$ OQ = OP + 3 = 8/3 + 3 = 17/3\$

10

Step

Now, let’s find the values of sin Q and cos Q:

\$ sin Q = (PQ)/(OQ) = 5/(17/3) = (15)/(17) \$

\$ cos Q = (OP)/(OQ) = (8/3)/(17/3) = 8/(17)\$

11

Step

Substitute the values:

\$ sin Q + cos Q = (15)/(17) + 8/(17) \$

\$ sin Q + cos Q = (23)/(17) \$

12

Solution

Therefore, the correct answer is \$ (23)/(17) \$.

13

Sumup

Please summarize Problem, Clue, Hint, Formula, Steps and Solution

Choices

14

Choice-A

This value is approximately 1.41, which isn’t consistent with the expected range of sine and cosine values (between -1 and 1)

Wrong \$( 24)/(17)\$

15

Choice-B

This value is approximately 1.03, which is closer to 1 but still outside the valid range for sine and cosine

Wrong \$( 41)/(40)\$

16

Choice-C

This value is greater than 1, which is not possible for the sum of sine and cosine in a right triangle

Wrong \$ (33)/(17)\$

17

Choice-D

This value falls within the acceptable range (-1 to 1) for the sum of sine and cosine in a right triangle

Correct \$ (23)/(17)\$

18

Answer

Option

D

19

Sumup

Please summarize choices