Question

The hypotenuse and one of the legs of a right triangle form an angle that has a sine of 4/9 . What is the tangent of the angle?

A) 4 (sq rt65)/65
B) 9(sq rt65)/65
C) (sq rt65)/4
D) (sq rt65)/9

Answer 1

Answer:

A

Step-by-step explanation:

So by drawing a right triangle, and knowing that Sin of an angle is the opposite over hypotenuse, we know what two of the three sides are. Using the Pythagorean theorem, we know that 4^2+b^2 = 9^2. Since 16+ b^2 = 81, we can solve for be to get b^2 = 65, which means that the missing leg is the square root of 65. Tan of an angle is opposite over adjacent, or 4/square root 65. since square roots can not be in the denominator, multiply by the fraction sq. rt. 65/ sq. rt. 65. After you multiply, you should get the answer of 4(sq.rt. 65)/ 65  

Answer 2

Answer:

(A) $tanC=\frac{4\sqrt{65}}{65}$

Step-by-step explanation:

Given: The hypotenuse and one of the legs of a right triangle form an angle that has a sine of ${\frac{4}{9}}$.

To find: The tangent of the angle.

Solution: It is given that  hypotenuse and one of the legs of a right triangle form an angle that has a sine of ${\frac{4}{9}}$, that is

AB=4 and AC=9

Now, using Pythagoras theorem, we have

$(AC)^2=(AB)^2+(BC)^2$

Substituting the given values, we get

$(9)^2=(4)^2+(BC)^2$

$81=16+(BC)^2$

$(BC)^2=65$

$(BC)=\sqrt{65}$

Now, using trigonometry, we have

$tanC=\frac{AB}{BC}$

Substituting the given values, we have

$tanC=\frac{4}{\sqrt{65}}$

$tanC=\frac{4\sqrt{65}}{65}$

Thus, option A is correct.