The product of 5 and x squared — 5(x)^2
Refer to c^2=a^2+b^2 (^2 means squared)
C is the hypotenuse and will be the longest measure
So 15^2= 9^2+12^2
225=225
If both sides are equal then the triangle is a right triangle, if the hypotenuse (or c) is lower or higher than you can determine whether it’s obtuse or acute
It would be a right triangle
Answer:
49.45
Step-by-step explanation:
because if u take 10.10 +20.15+19.20 = 49.45
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!
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! :)
recalling that d = rt, distance = rate * time.
we know Hector is going at 12 mph, and he has already covered 18 miles, how long has he been biking already?

so Hector has been biking for those 18 miles for 3/2 of an hour, namely and hour and a half already.
then Wanda kicks in, rolling like a lightning at 16mph.
let's say the "meet" at the same distance "d" at "t" hours after Wanda entered, so that means that Wanda has been traveling for "t" hours, but Hector has been traveling for "t + (3/2)" because he had been biking before Wanda.
the distance both have travelled is the same "d" miles, reason why they "meet", same distance.
![\bf \begin{array}{lcccl} &\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\ \cline{2-4}&\\ Hector&d&12&t+\frac{3}{2}\\[1em] Wanda&d&16&t \end{array}\qquad \implies \begin{cases} \boxed{d}=(12)\left( t+\frac{3}{2} \right)\\[1em] d=(16)(t) \end{cases}](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7Blcccl%7D%20%26%5Cstackrel%7Bmiles%7D%7Bdistance%7D%26%5Cstackrel%7Bmph%7D%7Brate%7D%26%5Cstackrel%7Bhours%7D%7Btime%7D%5C%5C%20%5Ccline%7B2-4%7D%26%5C%5C%20Hector%26d%2612%26t%2B%5Cfrac%7B3%7D%7B2%7D%5C%5C%5B1em%5D%20Wanda%26d%2616%26t%20%5Cend%7Barray%7D%5Cqquad%20%5Cimplies%20%5Cbegin%7Bcases%7D%20%5Cboxed%7Bd%7D%3D%2812%29%5Cleft%28%20t%2B%5Cfrac%7B3%7D%7B2%7D%20%5Cright%29%5C%5C%5B1em%5D%20d%3D%2816%29%28t%29%20%5Cend%7Bcases%7D)
