Sign In
Draw a right triangle where cos A= 47. Use the trigonometric ratio to label two side lengths and the Pythagorean Theorem to find the missing side.
sin A=sqrt(33)/7, tan A=sqrt(33)/4, csc A=7sqrt(33)/33, sec A=7/4, cot A=4sqrt(33)/33
In a right triangle, ∠A is an acute angle. Cosine is defined as the ratio of the adjecent side to the hypotenuse of an angle. We are given that cos A= 47.
cos = adj/hyp ⇒ cos A = 4 7
Now, let's draw a right triangle satisfying these conditions.
Let's find the opposite side by substituting b= 4 and c= 7 into the Pythagorean Theorem.
Note that when solving the equation we only kept the principal root. This is because a represents a side length, and therefore, must be positive.
Now that we know the length of all three sides of the triangle, we can find the values of the five remaining trigonometric functions. Remember to rationalize denominators if needed.
| Function | Substitute | Simplify |
|---|---|---|
| sin A=opp/hyp | sin A=sqrt(33)/7 | sin A=sqrt(33)/7 |
| tan A=opp/adj | tan A=sqrt(33)/4 | tan A=sqrt(33)/4 |
| csc A=hyp/opp | csc A=7/sqrt(33) | csc A=7sqrt(33)/33 |
| sec A=hyp/adj | sec A=7/4 | sec A=7/4 |
| cot A=adj/opp | cot A=4/sqrt(33) | cot A=4sqrt(33)/33 |
a/b=a * sqrt(33)/b * sqrt(33)
a* a=a^2
( sqrt(a) )^2 = a
Let's now follow the same procedure to rationalize the denominator of 4sqrt(33).
a/b=a * sqrt(33)/b * sqrt(33)
a* a=a^2
( sqrt(a) )^2 = a