HELPP HELP. It’s about slope!!!! Explanation needed
1 answer:
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Hi there! You have to remember these 6 basic Trigonometric Ratios which are:
- sine (sin) = opposite/hypotenuse
- cosine (cos) = adjacent/hypotenuse
- tangent (tan) = opposite/adjacent
- cosecant (cosec/csc) = hypotenuse/opposite
- secant (sec) = hypotenuse/adjacent
- cotangent (cot) = adjacent/opposite
- cosecant is the reciprocal of sine
- secant is the reciprocal of cosine
- cotangent is the reciprocal of tangent
Back to the question. Assuming that the question asks you to find the cosine, sine, cosecant and secant of angle theta.
What we have now are:
- Trigonometric Ratio
- Adjacent = 12
- Opposite = 10
Looks like we are missing the hypotenuse. Do you remember the Pythagorean Theorem? Recall it!
- a²+b² = c²
Define that c-term is the hypotenuse. a-term and b-term can be defined as adjacent or opposite
Since we know the value of adjacent and opposite, we can use the formula to find the hypotenuse.
- 10²+12² = c²
- 100+144 = c²
- 244 = c²
Thus, the hypotenuse is:
Now that we know all lengths of the triangle, we can find the ratio. Recall Trigonometric Ratio above! Therefore, the answers are:
- cosine (cosθ) = adjacent/hypotenuse = 12/(2√61) = 6/√61 = <u>(6√61) / 61</u>
- sine (sinθ) = opposite/hypotenuse = 10/(2√61) = 5/√61 = <u>(5√61) / 61</u>
- cosecant (cscθ) is reciprocal of sine (sinθ). Hence, cscθ = (2√61/10) = <u>√61/5</u>
- secant (secθ) is reciprocal of cosine (cosθ). Hence, secθ = (2√61)/12 = <u>√</u><u>61</u><u>/</u><u>6</u>
Questions can be asked through comment.
Furthermore, we can use Trigonometric Identity to find the hypotenuse instead of Pythagorean Theorem.
Hope this helps, and Happy Learning! :)
Answer:
The dimension of the box is l×w×h = (2.274 × 2.274 × 1.934) ft³
Step-by-step explanation:
From the given information;
Let a be the cost of the box
Let b be one side of the square base ; &
h to be the height of the box
We know that the volume of the box = 10 cubic feet
Then;
a²h = 10
h = 10/a²
The base = (0.65)a²
The top = (0.2)a²
The side = (0.2) a × 25/a²
= 5/a
For the four sides of the box now ;
= (0.2) 4a × 25/a²
= 0.8 × 25/a
= 20 /a
The total cost of the box is:
b = 0.65a² + 0.2a² + 20 /a
b = 0.85 a² + 20 /a
Taking differential of b with respect to a ;we have:
db/da = 1.7a - 1/a²(20) = 0
1.7 a³ - 20 = 0
1.7 a³ = 20
a³ = 20/1.7
a³ = 11.77
a =
a = 2.274 ft
Thus; the cost for the base of the box = (0.65)a²
the cost for the base of the box =(0.65) × ( 2.274)²
the cost for the base of the box = 3.362
The top of the box = (0.2)a²
The top of the box = (0.2)× ( 2.274)²
The top of the box = 1.034
The four sides of the box = 20 /a
The four sides of the box = 20/2.274
The four sides of the box = 8.795
the total cost = b = 0.85 a² + 20 /a
the total cost = 0.85 (2.274)² + 20 /2.274
the total cost = 4.395 + 8.795
the total cost = 13.19
Recall that:
the volume of the box = 10 cubic feet
Then;
a²h = 10
h = 10/a²
h = 10/ 2.274²
h 1.934
The dimension of the box is l×w×h = (2.274 × 2.274 × 1.934) ft³