Question

solve the right triangle.$sqrt{38}$$69^circ$write your answers as integers or as decimals rounded to the nearest tenth.$ik = square$$ij = square$$mangle k = square^circ$

Explanation:

Step1: Find $\angle K$

The sum of angles in a triangle is $180^\circ$. For right triangle $IJK$, $\angle I=90^\circ$, $\angle J=65^\circ$.
$m\angle K = 180^\circ - 90^\circ - 65^\circ = 25^\circ$

Step2: Calculate side $IK$

$IK$ is opposite $\angle J$, hypotenuse $KJ=\sqrt{38}$. Use $\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$.
$\sin(65^\circ)=\frac{IK}{\sqrt{38}}$
$IK = \sqrt{38} \times \sin(65^\circ) \approx 6.2 \times 0.9063 \approx 5.6$

Step3: Calculate side $IJ$

$IJ$ is adjacent to $\angle J$, hypotenuse $KJ=\sqrt{38}$. Use $\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$.
$\cos(65^\circ)=\frac{IJ}{\sqrt{38}}$
$IJ = \sqrt{38} \times \cos(65^\circ) \approx 6.2 \times 0.4226 \approx 2.6$

Answer:

$IK = 5.6$
$IJ = 2.6$
$m\angle K = 25^\circ$