When we find the ratio of two sides in a triangle, the ratio of the corresponding sides in a similar triangle will always be the same.

As such, this means that the trigonometric ratios (sine, cosine and tangent) in similar right-angle triangles are always equal.

**Example 1 :**

Find the value of csc E.

**Solution :**

In order to find csc E first we have to find the length all sides of triangle GEF.

By observing the above two given triangles, we can prove that the given triangles are similar.

The AA Similarity Theorem states that two triangles are similar if two angles of one triangle are congruent to two angles of the other triangle.

In triangles GEF and BCD,

∠G = ∠C (90 degee)

∠E = ∠B

Therefore, by the AA Similarity Theorem,

△DEF ~ △GIH

So, ratios of corresponding side lengths will be equal.

Then,

EF/BD = GF/CD = GE/BC -----(1)

In triangle EFG,

csc E = Hypotenuse side/ Opposite side

csc E = EF/GF -----(2)

From (2),

EF/BD = GF/CD

EF/GF = BD/CD

EF/GF = 37/35

Substitute 37/35 for EF/GF in (2).

(2)-----> csc E = 37/35

**Example 2 :**

Find the vale of sec D.

**Solution :**

By observing the above two triangles, we can prove that the given triangles are similar.

The AA Similarity Theorem states that two triangles are similar, if two angles of one triangle are congruent to two angles of the other triangle.

In triangles GHI and DFE,

∠I = ∠E (90 degree)

∠G = ∠D

Therefore, by the AA Similarity theorem,

△GHI ~ △DFE

So, the ratios of corresponding side lengths will be equal.

Then,

GH/DF = IH/EF = GI/DE -----(1)

sec D = Hypotenuse side/ Adjacent side

sec D = DF/DE -----(2)

From (1),

GH/DF = GI/DE

GH/GI = DF/DE

17/8 = DF/DE

Substitute 17/8 for DF/DE in (2).

(2)-----> sec D = 17/8

**Example 3 :**

Find the value of tan P.

**Solution :**

By observing the above two given triangles, we can prove that the given triangles are similar.

The AA Similarity Theorem states that two triangles are similar, if two angles of one triangle are congruent to two angles of the other triangle.

In triangles UST and PRQ,

∠S = ∠Q (90 degree)

∠T = ∠P

Therefore, by the AA Similarity theorem,

△UST ~ △PRQ

So, the ratios of corresponding side lengths will be equal.

Then,

UT/RP = US/RQ = TS/PQ -----(1)

tan P = Opposite side / Adjacent side

tan P = RQ/PQ -----(2)

From (1),

US/RQ = TS/PQ

US/TS = RQ/PQ

20/21 = RQ/PQ

Substitute 20/21 for RQ/PQ in (2).

(2)-----> tan P = 20/21

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