Question

trigonometric ratios of similar triangles
how can you use trigonometric ratios to show that these triangles are similar?

• we know that similar triangles will have congruent ______________ angles and

______________ side lengths.

• because of this property, _______ triangles will also have equal trigonometric ratios

for the corresponding _______ angles.
\\(\sin(q)=\frac{6}{3\sqrt{5}} = \frac{2}{\sqrt{\\_\\_}}\\) \\(\sin(l)=\frac{1}{\sqrt{5}}\\) \\(\sin(j)=\frac{2}{\sqrt{5}}\\)
\\(\cos(q)=\frac{3}{3\sqrt{5}} = \frac{1}{\sqrt{5}}\\) \\(\cos(l)=\frac{2}{\sqrt{\\_\\_}}\\) \\(\cos(j)=\frac{1}{\sqrt{5}}\\)
\\(\tan(q)=\frac{6}{3} = \\_\\_\\) \\(\tan(l)=\frac{1}{2}\\) \\(\tan(j) = 2\\)
(images of two right triangles: one with vertices o, p, q (right angle at p, op=6, pq=3, oq=3√5); another with vertices j, k, l (right angle at k, jk=1, kl=2, jl=√5))

• because the trigonometric ratios are not equal to each other, the angles cannot be

congruent. so \\(\angle q\\) does not correspond to \\(\angle l\\).

• \\(\angle q\\) is congruent to \\(\angle j\\) because their trigonometric ratios are ________.
• now that we have \\(\angle q\\) is congruent to \\(\angle j\\) and \\(\angle p\\) is congruent to \\(\angle k\\), because theyre

both right angles, then we can say that the triangles are similar by the aa similarity
theorem.
\\(\triangle qop \sim \triangle\\_\\_\\_\\)

Explanation:

Step1: Analyze Similar Triangles Property

Similar triangles have congruent corresponding angles and proportional side lengths. So first blank: "corresponding", second blank: "proportional".

Step2: Fill Trigonometric Ratios Blanks

For \(\sin(Q)=\frac{6}{3\sqrt{5}}=\frac{2}{\sqrt{5}}\), the denominator is \(\sqrt{5}\). For \(\cos(L)\), from triangle \(LJK\), hypotenuse is \(\sqrt{5}\), adjacent to \(L\) is \(2\)? Wait, no, triangle \(LJK\): \(JK = 1\), \(LK = 2\), \(LJ=\sqrt{5}\) (right triangle). So \(\cos(L)=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{2}{\sqrt{5}}\)? Wait, no, angle \(L\): adjacent side to \(L\) is \(LJ\)? Wait, no, right angle at \(J\), so angle \(L\): opposite is \(JK = 1\), adjacent is \(LJ\)? Wait, no, in \(\triangle LJK\), right-angled at \(J\), so sides: \(JK = 1\) (opposite \(L\)), \(LJ\) (adjacent to \(L\))? Wait, no, \(\cos(L)=\frac{\text{adjacent}}{\text{hypotenuse}}\). Hypotenuse \(LK=\sqrt{1^2 + 2^2}=\sqrt{5}\)? Wait, no, \(JK = 1\), \(LJ = 2\)? Wait, the diagram shows \(J\) to \(K\) is \(1\), \(K\) to \(L\) is \(2\), \(J\) to \(L\) is \(\sqrt{5}\). So right angle at \(J\), so angle \(L\): adjacent side is \(LJ = 2\), hypotenuse \(LK=\sqrt{5}\)? Wait, no, hypotenuse is \(LK\)? Wait, no, in right triangle, hypotenuse is opposite right angle. So right angle at \(J\), so hypotenuse is \(LK\), length \(\sqrt{1^2 + 2^2}=\sqrt{5}\). So \(\cos(L)=\frac{\text{adjacent to } L}{\text{hypotenuse}}=\frac{LJ}{LK}=\frac{2}{\sqrt{5}}\). So \(\cos(L)=\frac{2}{\sqrt{5}}\), so denominator is \(\sqrt{5}\). For \(\tan(Q)=\frac{6}{3}=2\).

Step3: Fill Congruent Angles and Similar Triangle

\(\angle Q\) and \(\angle J\) have equal trigonometric ratios (since \(\sin(Q)=\frac{2}{\sqrt{5}}\), \(\sin(J)=\frac{2}{\sqrt{5}}\); \(\cos(Q)=\frac{1}{\sqrt{5}}\), \(\cos(J)=\frac{1}{\sqrt{5}}\); \(\tan(Q)=2\), \(\tan(J)=2\)). So \(\angle Q\cong\angle J\), \(\angle P\cong\angle K\) (right angles), so by AA similarity, \(\triangle QOP \sim \triangle LKJ\) (since \(\angle Q\cong\angle J\), \(\angle P\cong\angle K\), so corresponding angles: \(Q\to J\), \(O\to L\), \(P\to K\)? Wait, no, \(\triangle QOP\) right-angled at \(P\), \(\triangle LJK\) right-angled at \(J\). So \(\angle P\cong\angle J\)? Wait, no, the problem says \(\angle P\cong\angle K\) (both right angles). Wait, \(\triangle QOP\): right at \(P\), \(\triangle LJK\): right at \(J\). Wait, maybe I mixed up. Wait, the text says " \(\angle P\) is congruent to \(\angle K\), because they're both right angles". So \(\angle P\) (right angle) \(\cong\angle K\) (right angle). Then \(\angle Q\cong\angle J\), so triangles \(\triangle QOP \sim \triangle LKJ\) (order: \(Q\to L\), \(O\to K\), \(P\to J\)? No, AA similarity: two angles congruent. \(\angle Q\cong\angle J\), \(\angle P\cong\angle K\), so \(\triangle QOP \sim \triangle LKJ\) (corresponding angles: \(Q\) and \(J\), \(P\) and \(K\), so \(O\) and \(L\)).

Step4: Fill Tan(Q)

\(\tan(Q)=\frac{\text{opposite}}{\text{adjacent}}=\frac{OP}{PQ}=\frac{6}{3}=2\).

Step5: Trigonometric Ratios Equality

\(\angle Q\) is congruent to \(\angle J\) because their trigonometric ratios are equal (since \(\sin(Q)=\frac{2}{\sqrt{5}}\), \(\sin(J)=\frac{2}{\sqrt{5}}\); \(\cos(Q)=\frac{1}{\sqrt{5}}\), \(\cos(J)=\frac{1}{\sqrt{5}}\); \(\tan(Q)=2\), \(\tan(J)=2\)).

Step6: Similar Triangle Name

Since \(\angle Q\cong\angle J\) and \(\angle P\cong\angle K\) (right angles), by AA similarity, \(\triangle QOP \sim \triangle LKJ\).

Answer:

1. We know that similar triangles will have congruent \(\boldsymbol{\text{corresponding}}\) angles and \(\boldsymbol{\text{proportional}}\) side lengths.
2. Because of this property, \(\boldsymbol{\text{similar}}\) triangles will also have equal trigonometric ratios for the corresponding \(\boldsymbol{\text{acute}}\) angles.
3. \(\sin(Q)=\frac{6}{3\sqrt{5}}=\frac{2}{\boldsymbol{\sqrt{5}}}\), \(\cos(L)=\frac{2}{\boldsymbol{\sqrt{5}}}\), \(\tan(Q)=\frac{6}{3}=\boldsymbol{2}\).
4. \(\angle Q\) is congruent to \(\angle J\) because their trigonometric ratios are \(\boldsymbol{\text{equal}}\).
5. \(\triangle QOP \sim \triangle \boldsymbol{LKJ}\)