Trigonometry Question

Question

Find the area of triangle ABC where a = 35, b = 26, and m/ C = 95º 52".  (Whoa! Did you notice the evil trick?)

ROUND TO THE NEAREST TENTH

Answer 1

Find the area of triangle ABC where a = 35, b = 26, and m/ C = 95º 52".  (Whoa! Did you notice the evil trick?)

Let's make it easier; first; label it correctly:

Sides: "a" shortest = 26

b mid 35

c hypotenuse

Angles: "α" opposite side "a".

"β" b

γ c. =95.866......° Changed into a decimal.

To get the area: bh/2.

To get side "c": a^2+b^2-2ab(cosγ)=c^2

26^2+35^2-(2*26*35)(cos95.866)=c^2

676 [phone]) * -.1022... =c^2

1901-(1820*(-.1022)=2087...^.5=45.6837...hypotenuse

We need angles α and β.

a/sinα=b/sinβ=c/sinγ

26/sinα=35/sinβ=45.683/sin96°=45/.9947=45.924;

26/45.924=.566;asin=34.48°=α

180-34.48-96=49.52°=β

γ=96°

This triangle is to be made into 2 rt triangles. Side "c" of the rt triangle is 26. Need side "a".

a/sin34.5=26 (90sin=1); a=.566*26=14.71

area=45.7*14.71/2=338area

Answer 2

c^2 = a^2 + b^2 - 2abcosC

c^2 = 35^2 + 26^2 - 2(35)26)cos95 52/60 degrees = 1225 + 676+182

c=sqr2073= 45.5

use Heron's formula

p=(35+26+45.5)/2 = 106.5/2= 53.3

Area = square root of ((p)(53.3-35)(53.3-26)(53.3-45.5))

= sqr(53.3)(18.3)(27)(7.8) = sqr(205,417.134)

= about a little over 453

cos90 = 0

cos95 or 96 is close to 0

c^2 = about 35^2 + 26^2

area should be = about 35 x26/2 = 35x13 = 455

It's almost a right triangle with c opposite the almost right angle