Sin A = 90-Cos A. And Cos A = 90-Sin A.
1 answer:
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The right triangles that have an altitude which forms two right triangles
are similar to the two right triangles formed.
Responses:
1. ΔLJK ~ ΔKJM
ΔLJK ~ ΔLKM
ΔKJM ~ ΔLKM
2. ΔYWZ ~ ΔZWX
ΔYWZ ~ ΔYZW
ΔZWX ~ ΔYZW
3. x = <u>4.8</u>
4. x ≈ <u>14.48</u>
5. x ≈ <u>11.37</u>
6. G.M. = <u>12·√3</u>
7. G.M. = <u>6·√5</u>
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<h3>What condition guarantees the similarity of the right triangles?</h3>1. ∠LMK = 90° given
∠JMK + ∠LMK = 180° linear pair angles
∠JMK = 180° - 90° = 90°
∠JKL ≅ ∠JMK All 90° angles are congruent
∠LJK ≅ ∠LJK reflexive property
- <u>ΔLJK is similar to ΔKJM</u> by Angle–Angle, AA, similarity postulate
∠JLK ≅ ∠JLK by reflexive property
- <u>ΔLJK is similar to ΔLKM</u> by AA similarity
By the property of equality for triangles that have equal interior angles, we have;
- <u>ΔKJM ~ ΔLKM</u>
2. ∠YWZ ≅ ∠YWZ by reflexive property
∠WXZ ≅ ∠YZW all 90° angle are congruent
- <u>ΔYWZ is similar to ΔZWX</u>, by AA similarity postulate
∠XYZ ≅ ∠WYZ by reflexive property
∠YXZ ≅ ∠YZW all 90° are congruent
- <u>ΔYWZ is similar to ΔYZW</u> by AA similarity postulate
Therefore;
- <u>ΔZWX ~ ΔYZW</u>
3. The ratio of corresponding sides in similar triangles are equal
From the similar triangles, we have;
8 × 6 = 10 × x
48 = 10·x
3. From the similar triangles, we have;
20 × 21 = x × 29
420 = 29·x
4. From the similar triangles, we have;
20 × 48 = 52 × x
5. From the similar triangles, we have;
13.2 × 22.4 = 26 × x
6. The geometric mean, G.M. is given by the formula;
The geometric mean of 16 and 27 is therefore;
- The geometric mean of 16 and 27 is <u>12·√3</u>
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7. The geometric mean of 5 and 36 is found as follows;
- The geometric mean of 5 and 36 is <u>6·√5</u>
Learn more about the AA similarity postulate and geometric mean here:
