52.7k views
0 votes
In ΔQRS, r = 510 cm, ∠S=5° and ∠Q=15°. Find the area of ΔQRS, to the nearest centimeter.

User Ksb
by
8.7k points

1 Answer

3 votes

Answer: The area is 8,652 square centimeters (rounded off to the nearest centimeter)

Step-by-step explanation: The triangle has one side given and two angles given as well. The third angle can be calculated as follows;

Angle R = 180 - (5 + 15)

Angle R = 180 - 20

Angle R = 160

To calculate side s we shall apply the Sine rule.

s/SinS = r/SinR

s/Sin5 = 510/Sin160

By cross multiplication we now have

s = (510 x Sin5)/Sin160

s = (510 x 0.0872)/0.3420

s = 44.472/0.3420

s = 130

Next we calculate side q as follows;

q/SinQ = r/SinR

q/Sin15 = 510/Sin160

By cross multiplication we now have

q = (510 x Sin15)/Sin160

q = (510 x 0.2588)/0.3420

q = 131.988/0.3420

q = 385.929

q = 386 (approximately)

Having found all three sides of the triangle as 130, 386 and 510 respectively the area shall be calculated by use of the Heron’s formula which is;

A = [square root] s(s - a)(s - b)(s - c)

Where s is the semi-perimeter, a, b and c are the three sides. The semi-perimeter is calculated as follows;

s = (130 + 386 + 510)/2

s = 1026/2

s = 513

The area now becomes

A = [square root] 513(513 - 510)(513 - 386)(513 - 130)

A = [square root] 513(3)(127)(383)

A = [square root] 74858499

A = 8652.0806

Rounded to the nearest centimeter, the area of the triangle is 8,652 square centimeters.

User Mvasco
by
8.7k points

Related questions

asked Feb 26, 2021 202k views
AntoG asked Feb 26, 2021
by AntoG
8.3k points
2 answers
2 votes
202k views