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Solve for X, round to the nearest tenth of a degree is necessary

Solve for X, round to the nearest tenth of a degree is necessary-example-1

1 Answer

4 votes

Answer:

x= 35.2344 °

Explanation:

Sides:

a = 2.6 m

b = 2.12368 m

c = 1.5 m

Angles:

A = 90 °

B = 54.7656 °

C = 35.2344 °

Other:

P = 6.22368 m

s = 3.11184 m

K = 1.59276 m2

r = 0.511838 m

R = 1.3 m

C = angle C

a = side a

b = side b

c = side c

P = perimeter

s = semi-perimeter

K = area

r = radius of inscribed circle

R = radius of circumscribed circle

AAS is Angle, Angle, Side

Given the size of 2 angles and 1 side opposite one of the given angles, you can calculate the sizes of the remaining 1 angle and 2 sides.

Use the Sum of Angles Rule to find the other angle, then

use The Law of Sines to solve for each of the other two sides.

User Yogesh Jilhawar
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