Question

1.4.3 quiz: solving problems with right triangles: mastery test solve the given right triangle for its missing angle and side measures. note: figure not drawn to scale a. m∠d = 55°,de≈8.40 units,df≈14.65 units b. m∠d = 35°,de≈8.40 units,df≈13.65 units c. m∠d = 55°,de≈4.40 units,df≈13.65 units d. m∠d = 35°,de≈8.40 units,df≈14.65 units

Explanation:

Step1: Find the missing angle

In a right - triangle, the sum of the interior angles is 180°. Given one non - right angle is 35° and the right angle is 90°. Let the missing angle be $\angle D$. Then $\angle D=180^{\circ}-90^{\circ}-35^{\circ}=55^{\circ}$.

Step2: Find side $DE$

We know that $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$. Here, $\theta = 35^{\circ}$ and the adjacent side to $\angle F$ is $EF = 12$ units, and the opposite side is $DE$. So, $\tan35^{\circ}=\frac{DE}{12}$, and $DE = 12\times\tan35^{\circ}\approx12\times0.7002\approx8.40$ units.

Step3: Find side $DF$

We know that $\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$. Here, $\theta = 35^{\circ}$, the adjacent side to $\angle F$ is $EF = 12$ units, and the hypotenuse is $DF$. So, $\cos35^{\circ}=\frac{12}{DF}$, and $DF=\frac{12}{\cos35^{\circ}}\approx\frac{12}{0.8192}\approx14.65$ units.

Answer:

A. $m\angle D = 55^{\circ},DE\approx8.40$ units,$DF\approx14.65$ units