Table of Contents

Last modified on August 3rd, 2023

SAS means ‘Angle-Side-Angle’. SAS triangles are triangles where two sides and the angle between them are known. Shown below is a SAS triangle, △ABC, where b and c are the two known sides, while ∠B is the known, common angle present between the two sides.

It involves three steps:

**Step 1**: Use the Law of Cosines to calculate the missing side

**Step 2**: Use The Law of Sines to find the smaller of the two unknown angles

**Step 3**: Use the angle sum rule of a triangle to find the last angle

Let us take some examples to understand the concept better.

**Find the missing angles and the side in the given SAS triangle.**

Solution:

In the triangle, the given angles and side is:

Side b = 4

Side c = 6

∠A = 45°**Step 1:**

Using theLaw of Cosines, we will calculate the missing side, side a

a^{2} = b^{2} + c^{2} − 2bc cos A, here b = 4, c = 6, ∠A = 45°

a^{2} = 4^{2} + 6^{2} – 2 x 4 x 6 x cos 45°

a^{2} = 16 + 36 – 48 x cos 45°

a^{2} = 52 – 48 x 0.70

a^{2} = 52 – 33.6

a^{2} = 18.4

a = 4.28**Step 2:**

Now, we use The Law of Sines to find the smaller of the two unknown angles**Why the Smaller Angle to be Determined First?**

The smaller angle is determined first because the inverse sine function gives answers less than 90° even for angles greater than 90°. By choosing the smaller angle a triangle cannot have two angles greater than 90°.

a/sin A = b/sin B, here ∠A = 45°, a = 4.28, b = 4

4.28/sin 45° = 4/sin B

sin B = 4 x sin 45°/4.28

sin B = 4 x 0.70/4.28

sin B = 0.65

B = sin ^{-1} (0.65)

B = 40.54°**Step 3:**

Finally, we will use the angle sum rule of a triangle to find the last undetermined angle, ∠C

∠A + ∠B + ∠C = 180°, here ∠A = 45°, ∠B = 40.54°

45° + 40.54° +∠C = 180°

∠C = 180° – (45° + 40.54°)

∠C = 94.46°

**Find the missing angles and the side in the given triangle.**

Solution:

The given triangle is an SAS triangle.

Here,

Side a = 6.8

Side b = 4.2

∠C = 110°**Step 1:**

Using theLaw of Cosines, we will calculate the missing side, side c

c^{2} = a^{2} + b^{2} − 2ab cos C, here a = 6.8, b = 4.2, ∠C = 110°

c^{2} = 6.8^{2} + 4.2^{2} – 2 x 6.8 x 4.2 x cos 110°

c^{2} = 46.24 + 17.64 – 57.12 x (-0.34)

c^{2} = 63.88 + 19.42

c^{2} = 83.3

c = 9.12**Step 2:**

Now, for calculating the Law of Sines, we don’t need to look for the smaller angle as ∠C is greater than 90°, so ∠A and ∠B must be less than 90°.

a/sin A = c/sin C, here, a = 6.8, c = 9.12, ∠C = 110°

6.8/sin A = 9.12/ sin 110°

sin A = 6.8 x sin 110°/9.12

sin A = 6.8 x 0.93/9.12

sin A = 0.69

A = sin^{-1} (0.69)

∠A = 43.63°**Step 3:**

Finally, we will use the angle sum rule of a triangle to find the last undetermined angle, ∠B

∠A + ∠B + ∠C = 180°, here ∠A = 43.63°, ∠C = 110°

43.63° + ∠B + 110° = 180°

∠B = 180° – (43.63° + 110°)

∠B = 26.37°

Let us take an example to understand the concept.

**To prove:** **ΔABO ****≅ ΔCDO**

**Proof**:

Steps | Statements | Reasons |

1. | AB ≅ CD AB || CD | Given |

2. | ∠BAO ≅∠DCO | Alternate interior angles |

3. | ∠ABO ≅∠CDO | Alternate interior angles |

4. | ΔABO ≅ ΔCDO | SAS postulate (Hence proved) |

**Identify which pair of triangles below illustrates an SAS relationship to prove the triangles are congruent.**

Solution:

Option (b). Here the triangles ΔHIJ and ΔKLM are congruent by the SAS relationship.

**Prove whether ΔABC and ΔDEC are congruent by SAS. Given that, BC ≅ DC and C is the midpoint of AE.**

Solution:

Given that BC ≅ DC and C is the midpoint of AE

Now, according to the ‘Vertical Angles Theorem‘ ∠ACB ≅∠ECD

Again, AC ≅ EC as the midpoint C bisects AE into two congruent parts AC and EC

Hence **ΔABC ≅ ΔDEC** by SAS postulate (Proved)

If ΔABC is an SAS triangle with side lengths a, b and c, then we can find its area with one of the three formulas given below:

The method discussed above is just an extension of the general formula for determining the area of a triangle.

As we know, the general formula of area of a triangle is given by,

Area (A) = ½ x a x h, here a = base, h = height

But, if we are not given the height but we are given side b and angle C in the right triangle, ΔCDA

Using trigonometry, we can write

sin C = h/b

=> h = b sin C

Substituting the value of h in the general formula of area of a triangle, we get,

Area (A) = ½ x a x b sin C = ½ ab sin C

Let us solve an example using the above formula.

**Find the area of the given triangle.**

Solution:

As the given triangle is an SAS triangle, we will use the formula

Area (A) = ½ ab sin C, here a = 10, b = 16, ∠C = 55°

= ½ 10 x 16 sin 55°

= 64.8 square units

Last modified on August 3rd, 2023