Question

multiple choice
identify the choice that best completes the statement or answers the question.

1. determine the measure of ∠d to the nearest tenth of a degree.

triangle diagram with right angle at e, de=14, ep=8
a. 29.7° b. 34.8° c. 55.2° d. 60.3°

1. determine tan q and tan r.

triangle diagram with right angle at p, pq=6, pr=10
a. tan q = 0.375; tan r = 0.6 c. tan q = 0.6; tan r = 1.6
b. tan q = 1.6; tan r = 0.6 d. tan q = 1.6; tan r = 0.625

1. determine tan a and tan c.

triangle diagram with right angle at b, ab=6, bc=4
a. tan a = 0.6; tan c = 1.5 c. tan a = 1.5; tan c = 0.6
b. tan a = 0.5547…; tan c = 1.5 d. tan a = 0.6; tan c = 0.8321…

Explanation:

Question 1

Step1: Identify opposite and adjacent sides

In the right triangle, for $\angle D$, the opposite side to $\angle D$ is $8$ and the adjacent side is $14$.

Step2: Use arctangent formula

We know that $\tan(\angle D)=\frac{\text{opposite}}{\text{adjacent}}=\frac{8}{14}\approx0.5714$. Then $\angle D = \arctan(0.5714)$.

Step3: Calculate the angle

Using a calculator, $\arctan(0.5714)\approx29.7^{\circ}$ (wait, no, wait: Wait, maybe I mixed up opposite and adjacent. Wait, looking at the triangle, $DE = 14$ (adjacent to $\angle D$), $EP = 8$ (opposite to $\angle D$)? Wait, no, maybe the right angle is at $E$, so triangle $DEP$ with right angle at $E$. So $\angle D$: opposite side is $EP = 8$, adjacent side is $DE = 14$. So $\tan(\angle D)=\frac{8}{14}\approx0.5714$, so $\angle D=\arctan(0.5714)\approx29.7^{\circ}$? But wait, the options are a. $29.7^{\circ}$, b. $34.8^{\circ}$, c. $55.2^{\circ}$, d. $60.3^{\circ}$. Wait, maybe I got opposite and adjacent wrong. Wait, maybe the vertical side is $8$, horizontal is $14$. So $\tan(\angle D)=\frac{8}{14}\approx0.571$, arctan(0.571) is about $29.7^{\circ}$, so option a? Wait, no, wait, maybe the other angle. Wait, the angle at $D$: if we consider the right triangle, the two acute angles add up to $90^{\circ}$. Wait, maybe I made a mistake. Wait, let's recalculate: $\tan(\theta)=\frac{8}{14}\approx0.571$, $\theta=\arctan(0.571)\approx29.7^{\circ}$, so option a.

Step1: Recall tangent definition

In a right triangle, $\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}$.

Step2: For $\tan Q$

In triangle $PQR$ (right-angled at $P$), for $\angle Q$: opposite side is $PR = 10$, adjacent side is $PQ = 6$. So $\tan Q=\frac{10}{6}=\frac{5}{3}\approx1.666... = 1.\overline{6}$.

Step3: For $\tan R$

For $\angle R$: opposite side is $PQ = 6$, adjacent side is $PR = 10$. So $\tan R=\frac{6}{10}=0.6$.

So $\tan Q = 1.\overline{6}$ and $\tan R = 0.6$, which is option b.

Step1: Recall tangent definition

In right triangle $ABC$ (right-angled at $B$), $\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}$.

Step2: For $\tan A$

For $\angle A$: opposite side is $BC = 4$, adjacent side is $AB = 6$. Wait, no: Wait, $AB = 6$, $BC = 4$. So $\tan A=\frac{BC}{AB}=\frac{4}{6}=\frac{2}{3}\approx0.666... = 0.\overline{6}$? Wait, no, wait the options: option a: $\tan A = 0.\overline{6}$; $\tan C = 1.5$, option c: $\tan A = 1.5$; $\tan C = 0.\overline{6}$. Wait, maybe I mixed up. Wait, $\angle A$: opposite side is $BC = 4$, adjacent is $AB = 6$? No, wait $AB$ is horizontal, $BC$ is vertical. So $\angle A$: opposite is $BC = 4$, adjacent is $AB = 6$? Then $\tan A=\frac{4}{6}=\frac{2}{3}\approx0.666$, but option a is $\tan A = 0.\overline{6}$ (which is $\frac{2}{3}\approx0.666$) and $\tan C = 1.5$. Wait, $\angle C$: opposite side is $AB = 6$, adjacent side is $BC = 4$. So $\tan C=\frac{6}{4}=1.5$. So $\tan A=\frac{4}{6}=0.\overline{6}$, $\tan C=\frac{6}{4}=1.5$, which is option a? Wait, no, option a: $\tan A = 0.\overline{6}$; $\tan C = 1.5$. Wait, but let's check the options again. Option a: $\tan A = 0.\overline{6}$; $\tan C = 1.5$. Option c: $\tan A = 1.5$; $\tan C = 0.\overline{6}$. Wait, I think I made a mistake. Wait, $\angle A$: opposite side is $BC = 4$, adjacent is $AB = 6$? No, wait $AB$ is the adjacent side to $\angle A$, and $BC$ is the opposite side. So $\tan A=\frac{\text{opposite}}{\text{adjacent}}=\frac{BC}{AB}=\frac{4}{6}=\frac{2}{3}\approx0.666 = 0.\overline{6}$. $\angle C$: opposite side is $AB = 6$, adjacent side is $BC = 4$. So $\tan C=\frac{AB}{BC}=\frac{6}{4}=1.5$. So that's option a? Wait, but let's check the options:

a. $\tan A = 0.\overline{6}$; $\tan C = 1.5$

b. $\tan A = 0.5547...$; $\tan C = 1.5$

c. $\tan A = 1.5$; $\tan C = 0.\overline{6}$

d. $\tan A = 0.\overline{6}$; $\tan C = 0.8321...$

Wait, so according to this, option a is correct? Wait, no, wait I think I mixed up opposite and adjacent. Wait, $\angle A$: in triangle $ABC$, right-angled at $B$, so sides: $AB = 6$ (horizontal), $BC = 4$ (vertical), $AC$ (hypotenuse). So $\angle A$: the angle at $A$, so the sides: opposite to $\angle A$ is $BC = 4$, adjacent to $\angle A$ is $AB = 6$. So $\tan A=\frac{BC}{AB}=\frac{4}{6}=\frac{2}{3}\approx0.666 = 0.\overline{6}$. $\angle C$: angle at $C$, opposite side is $AB = 6$, adjacent side is $BC = 4$. So $\tan C=\frac{AB}{BC}=\frac{6}{4}=1.5$. So yes, option a. Wait, but let's check the options again. Option a: $\tan A = 0.\overline{6}$; $\tan C = 1.5$. So that's correct.

Answer:

a. $29.7^{\circ}$

Question 2