El triple del área de un cuadrado menos seis veces la medida de su lado es igual a 0. ¿Cuánto mide por lado el cuadrado?
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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! :)