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find the area of a triangle when a=22 degree, b=105 degree, and b=14 30.4 units^2 32.1 units^2 31.2 units^2 34.8 units^2

2 Answers

2 votes
C = 180 - (22 + 105) = 180 - 127 = 53 degrees
Area of triangle = 1/2 ab sin C
But a / sin A = b / sin B or a = b sin A / sin B = (14 x sin 22) / sin 105 = 5.429
Therefore, Area of triangle = 1/2 x 5.429 x 14 x sin 53 = 30.4 units^2
User Gfoidl
by
7.3k points
3 votes

Answer:

Option A. 30.4 unit²

Explanation:

In the given triangle ABC, side AC = 14 units, ∠ BAC = 22° and ∠ ABC = 105°.

We have to find the area of ΔABC.

Since area of an obtuse triangle =
(1)/(2)(Base)(Height)

Here Base = x

Height = h

To find the measure of Base "x".

We will apply the sine rule in ΔABC


(sin22)/(x)=(sin105)/(y14)


(0.375)/(x)=(0.966)/(14)

We cross multiply on both the sides

14×0.375 = 0.966×(x)

5.25 = 0.966x

x =
(5.25)/(0.966)

x = 5.43 unit

For the measurement of Height "h"


cos(15+22)=(Base)/(Hypotenuse)


cos37=(h)/(14)


0.978=(h)/(14)

h = 0.978×14 = 11.18 unit

Now we put the values of base and height in the formula

Area of Δ ABC =
(1)/(2)(5.43)(11.18)

=
(60.71)/(2)

= 30.35 ≈ 30.4 unit²

find the area of a triangle when a=22 degree, b=105 degree, and b=14 30.4 units^2 32.1 units-example-1
User Olle Ekberg
by
6.9k points
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