SaiBook.us SAT-2

Home Course Section

Schedule Admission Social

Lesson Example Discussion Quiz: Class Homework

Step-1

Title: Trigonometry ratios in right triangles

Grade: 1400-a Lesson: S3-L2

Explanation: Hello Students, time to practice and review the steps for the problem.

Lesson Steps

$1$

Step	Type	Explanation	Answer
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 = 5cm, OQ - OP = 3cm
4	Hint	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 + 6"OP" = "OP"^2 + 25 \$ \$ 9 + 6"OP" = 25 \$
8	Step	After simplification	\$ 6"OP" = 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	Step	Therefore, the correct answer is \$ 23/17 \$.
13	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)	\$ 24/17\$
14	Choice.B	This value is approximately 1.03, which is closer to 1 but still outside the valid range for sine and cosine	\$ 41/40\$
15	Choice.C	This value is greater than 1, which is not possible for the sum of sine and cosine in a right triangle	\$ 33/17\$
16	Choice.D	This value falls within the acceptable range (-1 to 1) for the sum of sine and cosine in a right triangle	\$ 23/17\$
17	Answer	Option	D
18	Sumup	Can you summarize what you’ve understood in the above steps?

Step

Type

Explanation

Answer

Problem

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

Step

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

Step

The given values are

PQ = 5cm, OQ - OP = 3cm

Hint

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 \$

Step

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

OQ = OP + 3

Step

Now, substitute this into the Pythagorean theorem:

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

Step

Expand and simplify the equation:

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

Step

After simplification

\$ 6"OP" = 16 \$
\$ "OP" = 16/6 = 8/3 \$

Step

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

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

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\$

Step

Substitute the values:

\$ "sin Q" + "cos Q" = 15/17 + 8/17 \$

\$ "sin Q" + "cos Q" = 23/17 \$

Step

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

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)

\$ 24/17\$

Choice.B

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

\$ 41/40\$

Choice.C

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

\$ 33/17\$

Choice.D

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

\$ 23/17\$

Answer

Option

Sumup

Can you summarize what you’ve understood in the above steps?