Question

1. in each triangle, write the tangent ratio for each acute angle.

a)
triangle abc with right angle at b, bc=6, ab=7
b)
triangle def with right angle at e, de=4, ef=6
c)
triangle ghj with right angle at g, gj=10, gh=8
d)
triangle kmn with right angle at n, kn=7, mn=5

Explanation:

Part (a)

Step1: Recall tangent ratio

In a right triangle, \(\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}\) for an acute angle \(\theta\). For \(\triangle ABC\) with right angle at \(B\), we have two acute angles: \(\angle A\) and \(\angle C\).

Step2: Tangent of \(\angle A\)

For \(\angle A\), opposite side is \(BC = 6\), adjacent side is \(AB=7\). So \(\tan(A)=\frac{BC}{AB}=\frac{6}{7}\).

Step3: Tangent of \(\angle C\)

For \(\angle C\), opposite side is \(AB = 7\), adjacent side is \(BC = 6\). So \(\tan(C)=\frac{AB}{BC}=\frac{7}{6}\).

Part (b)

Step1: Recall tangent ratio

In \(\triangle DEF\) with right angle at \(E\), acute angles are \(\angle D\) and \(\angle F\).

Step2: Tangent of \(\angle D\)

For \(\angle D\), opposite side is \(EF = 6\), adjacent side is \(DE = 4\). So \(\tan(D)=\frac{EF}{DE}=\frac{6}{4}=\frac{3}{2}\).

Step3: Tangent of \(\angle F\)

For \(\angle F\), opposite side is \(DE = 4\), adjacent side is \(EF = 6\). So \(\tan(F)=\frac{DE}{EF}=\frac{4}{6}=\frac{2}{3}\).

Part (c)

Step1: Recall tangent ratio

In \(\triangle GHJ\) with right angle at \(G\), acute angles are \(\angle J\) and \(\angle H\).

Step2: Tangent of \(\angle J\)

For \(\angle J\), opposite side is \(GH = 8\), adjacent side is \(GJ = 10\). So \(\tan(J)=\frac{GH}{GJ}=\frac{8}{10}=\frac{4}{5}\).

Step3: Tangent of \(\angle H\)

For \(\angle H\), opposite side is \(GJ = 10\), adjacent side is \(GH = 8\). So \(\tan(H)=\frac{GJ}{GH}=\frac{10}{8}=\frac{5}{4}\).

Part (d)

Answer:

s:

(a)

\(\tan(A)=\frac{6}{7}\), \(\tan(C)=\frac{7}{6}\)

(b)

\(\tan(D)=\frac{3}{2}\), \(\tan(F)=\frac{2}{3}\)

(c)

\(\tan(J)=\frac{4}{5}\), \(\tan(H)=\frac{5}{4}\)

(d)

\(\tan(K)=\frac{5}{7}\), \(\tan(M)=\frac{7}{5}\)