When working with an SAS triangle—a triangle where two sides and the included angle are known—it is essential to follow a structured approach. The Side-Angle-Side (SAS) method involves first finding the unknown side and then determining the other two angles.
Solving SAS Triangles
For example, consider a triangle where Angle A = 49°, side b = 5, and side c = 7. The first step is to use the Law of Cosines to calculate the missing side a:
a2=b2+c2−2bc⋅cos(A)a² = b² + c² – 2bc \cdot \cos(A)a2=b2+c2−2bc⋅cos(A)
By substituting the given values:
a2=52+72−2×5×7×cos(49°)a² = 5² + 7² – 2 \times 5 \times 7 \times \cos(49°)a2=52+72−2×5×7×cos(49°)
Solving this equation step by step:
· 52=255² = 2552=25, 72=497² = 4972=49
· 2×5×7=702 \times 5 \times 7 = 702×5×7=70
· cos(49°)≈0.6561\cos(49°) \approx 0.6561cos(49°)≈0.6561
a2=25+49−(70×0.6561)=28.075a² = 25 + 49 – (70 \times 0.6561) = 28.075a2=25+49−(70×0.6561)=28.075
Taking the square root,
a≈5.30a ≈ 5.30a≈5.30
(rounded to two decimal places).
Now that side a is found, we proceed to determine the other two angles.
Using the Law of Sines to Find the Smaller Angle
The Law of Sines states:
sin(A)a=sin(B)b=sin(C)c\frac{\sin(A)}{a} = \frac{\sin(B)}{b} = \frac{\sin(C)}{c}asin(A) =bsin(B) =csin(C)
For Angle B:
sin(B)=b⋅sin(A)a\sin(B) = \frac{b \cdot \sin(A)}{a}sin(B)=ab⋅sin(A)
Substituting values:
sin(B)=5×sin(49°)5.298\sin(B) = \frac{5 \times \sin(49°)}{5.298}sin(B)=5.2985×sin(49°)
Since sin(49°) ≈ 0.7547:
sin(B)=5×0.75475.298=0.7122\sin(B) = \frac{5 \times 0.7547}{5.298} = 0.7122sin(B)=5.2985×0.7547 =0.7122
Applying the inverse sine function (sin⁻¹):
B≈45.4°(rounded to one decimal place)B ≈ 45.4° \quad \text{(rounded to one decimal place)}B≈45.4°(rounded to one decimal place)
Finding Angle C
Since the three angles in a triangle must add up to 180°:
C=180°−(49°+45.4°)=85.6°C = 180° – (49° + 45.4°) = 85.6°C=180°−(49°+45.4°)=85.6°
Thus, Angle C ≈ 85.6° (rounded to one decimal place).
Example 2
Consider another SAS triangle with:
· p = 6.9, q = 2.6, R = 117°
To find side r, use the Law of Cosines:
r2=p2+q2−2pq⋅cos(R)r² = p² + q² – 2pq \cdot \cos(R)r2=p2+q2−2pq⋅cos(R)
Substituting the values:
r2=(6.9)2+(2.6)2−2×6.9×2.6×cos(117°)r² = (6.9)² + (2.6)² – 2 \times 6.9 \times 2.6 \times \cos(117°)r2=(6.9)2+(2.6)2−2×6.9×2.6×cos(117°) r2=47.61+6.76−(35.88×(−0.4535))r² = 47.61 + 6.76 – (35.88 \times (-0.4535))r2=47.61+6.76−(35.88×(−0.4535)) r2=47.61+6.76+18.36=70.659r² = 47.61 + 6.76 + 18.36 = 70.659r2=47.61+6.76+18.36=70.659
Taking the square root:
r≈8.41r ≈ 8.41r≈8.41
(rounded to two decimal places).
Now, applying the Law of Sines to find Angle P:
sin(P)=p⋅sin(R)r\sin(P) = \frac{p \cdot \sin(R)}{r}sin(P)=rp⋅sin(R)
Substituting values:
sin(P)=6.9×sin(117°)8.41\sin(P) = \frac{6.9 \times \sin(117°)}{8.41}sin(P)=8.416.9×sin(117°)
Since sin(117°) ≈ 0.8910:
sin(P)=6.9×0.89108.41=0.7313\sin(P) = \frac{6.9 \times 0.8910}{8.41} = 0.7313sin(P)=8.416.9×0.8910 =0.7313
Taking the inverse sine (sin⁻¹):
Since the three angles in a triangle must add up to 180°:
C=180°−(49°+45.4°)=85.6°C = 180° – (49° + 45.4°) = 85.6°C=180°−(49°+45.4°)=85.6°
Thus, Angle C ≈ 85.6° (rounded to one decimal place).
Example 2
Consider another SAS triangle with:
· p = 6.9, q = 2.6, R = 117°
To find side r, use the Law of Cosines:
r2=p2+q2−2pq⋅cos(R)r² = p² + q² – 2pq \cdot \cos(R)r2=p2+q2−2pq⋅cos(R)
Substituting the values:
r2=(6.9)2+(2.6)2−2×6.9×2.6×cos(117°)r² = (6.9)² + (2.6)² – 2 \times 6.9 \times 2.6 \times \cos(117°)r2=(6.9)2+(2.6)2−2×6.9×2.6×cos(117°) r2=47.61+6.76−(35.88×(−0.4535))r² = 47.61 + 6.76 – (35.88 \times (-0.4535))r2=47.61+6.76−(35.88×(−0.4535)) r2=47.61+6.76+18.36=70.659r² = 47.61 + 6.76 + 18.36 = 70.659r2=47.61+6.76+18.36=70.659
Taking the square root:
r≈8.41r ≈ 8.41r≈8.41
(rounded to two decimal places).
Now, applying the Law of Sines to find Angle P:
sin(P)=p⋅sin(R)r\sin(P) = \frac{p \cdot \sin(R)}{r}sin(P)=rp⋅sin(R)
Substituting values:
sin(P)=6.9×sin(117°)8.41\sin(P) = \frac{6.9 \times \sin(117°)}{8.41}sin(P)=8.416.9×sin(117°)
Since sin(117°) ≈ 0.8910:
sin(P)=6.9×0.89108.41=0.7313\sin(P) = \frac{6.9 \times 0.8910}{8.41} = 0.7313sin(P)=8.416.9×0.8910 =0.7313
Taking the inverse sine (sin⁻¹):
P≈47.0°P ≈ 47.0°P≈47.0°
Finally, Angle Q:
Q=180°−(117°+47.0°)=16.0°Q = 180° – (117° + 47.0°) = 16.0°Q=180°−(117°+47.0°)=16.0°
Thus, Angle Q ≈ 16.0°.
Application: Stadium Example
Consider a stadium designed as a regular pentagon, where each side is 920 feet. The distance from the center to the corners forms isosceles triangles, dividing the pentagon into 5 equal sections.
Each angle at the center is:
360°5=72°\frac{360°}{5} = 72°5360° =72°
Using the Law of Cosines, the side connecting two corners is calculated as:
approximately 782 feet or 783 feet.\text{approximately } 782 \text{ feet} \text{ or } 783 \text{ feet}.approximately 782 feet or 783 feet.
Area of a Triangle with 2 Sides and an Included Angle
The area of an SAS triangle is found using the formula:
Area=12bcsin(A)\text{Area} = \frac{1}{2} bc \sin(A)Area=21 bcsin(A)
For ∆ABC, where AB = 5, BC = 8, and ∠ABC = 60°:
Area=12×5×8×sin(60°)\text{Area} = \frac{1}{2} \times 5 \times 8 \times \sin(60°)Area=21 ×5×8×sin(60°)
Since sin(60°) = √3/2:
Area=12×5×8×32=103\text{Area} = \frac{1}{2} \times 5 \times 8 \times \frac{\sqrt{3}}{2} = 10\sqrt{3}Area=21 ×5×8×23 =103
Thus, the area is 10√3 square units.
For a quadrilateral BCED, we break it into two triangles. Given AC = 30 and ∠BAC = 20°, the area of ∆ADE can be found similarly.
By following this method, we can systematically solve SAS triangles in various real-world applications, ensuring accuracy through structured calculations.