a triangle has angles 35 degress and 65 degrees. if the shortest side is 30 yd, find the area of the triangle.

Question

you can use the law of sins in finding the length of the other sides..

Answer 1

Angles are A =35° , B = 65° , C = 80°
Side a = 30 yd
b/sin B = a/sinA
b = a sin B / sin A
b = 30 sin 65° / sin 35°
b = 47.4 yd
Area = (1/2) a b sin C
Area = (1/2) (30) (47.4) sin 80°
Area = 700 yd ² (to nearest whole number)

Answer 2

Using the law of sines, the side length opposite the 65 degree angle will be 47.4, and the side length opposite the 80 degree angle (180 - 65 - 35) will be 51.51.

Split the triangle into two right angle triangles.

On the left side, you have a 35 degree angle, and a hypotenuse of 51.51. Find the base and height:

cos 35 = b / 51.51
b = 42.1945

sin 35 = h / 51.51
h = 29.5449

A = bh / 2
A = 623.316

Now the other triangle. You have an angle of 80 degrees, and a hypotenuse of 30.

Find b and h

cos 80 = b / 30
b = 5.20945

sin 80 = h / 30
h = 29.54423

A = bh /2
A = 76.9546

Total A = A1 + A2 = 700.2706 square yards.

Answer 3

Given:
A = 35 deg
B = 65 deg
Therefore,
C = 180 - (35+65)
C = 80 deg
Shortest side = 30 yd
Therefore, A is opposite the shortest side.
Using Sine Law to solve for the other sides of the triangle:
a/SinA = b/SinB
30/Sin35 = b/Sin65
b = Sin65(30/Sin35)
b = 47.4 yd

a/SinA = c/SinC
30/Sin35 = c/Sin80
c = Sin80(30/Sin35)
c = 51.5 yd

Using Hero's Formula:
A = sqrt[s(s-a)(s-b)(s-c)]

where s = (1/2)(a + b + c) = (1/2)(30 + 47.4 + 51.5)
s = 64.45

A = sqrt[64.45(64.45-30)(64.45-47.4)(64.45-51.5)]
A = 700.2 yd^2 ANS

Hope I help you.

teddy boy

Answer 4

Now shortest length is opposite to the shortest angle .

sin rule :-

30/sin35 = X/sin65

x=47.40298221 yd

Third angle = 180 - 35 -65 = 80



Area of triangle = 0.5absin(t)
t=80
a=30
b=54.37846722
Area = 0.5(30)(47.40298221)sin80
Area = 700.242366 yd²

Answer 5

let y=side opposite 65degrees
now use sine law: y/sin65 = 30/sin35 to find y.

Now if you have y,

you can find the altitude of the triangle using sine law again. (use <65 and <80 as base angles)

let h=altitude

use h/sin80 = y/sin 90 to find the altitude h.

Now you have the altitude, and the value of the base is given which is the shortest side.

Use A=(bh)/2 to find the area of the triangle

A= {30yd*[30yd(sin65)(sin80)]/[(sin35)(sin90)]}/2

A=700.24yd^2

try solving it again. I just used an old school scientific calculator to solve taht.

Answer 6

let in Δ ABC, A = 35 deg B = 65 deg then C = 180 -- 65 -- 35 = 80 deg
then area of Δ ABC
= 1/2 bcsinA
= 0.5(30)^2sin65sin80sin35 / sin^2(35)
= 0.5(900)(0.9063)(0.9848)(0.5736) / (0.5736)^2 sq yds
= 700.2021 sq yds.