the length of the base of an isoceles triangle is 30m.The angle opposite base is 32degree Find the perimeter?

Answer 1

Draw a triangle ABC as follows:-
Base is BC = 30
A = 32°
B = C = 74°

Sine Rule
30 / sin 32° = AC / sin 74°
AC = 30 sin 74° / sin 32°
AC = 54.4
AB = 54.4

P = 108.8 + 30 m
P = 138.8 m

Answer 2

The perimeter, p , for an isoceles triangle is given by the forumla:

p = 2e + b

where e is the length of the matching edges,
and b is the length of the base.

In your case, we know the base... b = 30


p = 2e + 30

Therefore, this problem reduces to finding the value of "e". If we find "e", we know the answer.

You can easily form 2 right triangles in an isoceles triangle if you bisect the base with a perpendicular line segment, which should strike thorugh where the 2 "e" pieces touch.

That gives you a triangle with hypotenuse = e, base, b/2, and unknown distance, h.

we do not know the value of h, but we do know that b/2, or half the base, is 30/2 = 15.

Since we know the angle between the base and "e" is 32° degrees, we can calculate using trignometric identities:

cos 32° = 15/e , therefore e = 15 / cos 32° = 17.68

and thus... p = 2e + b =

p = 2*17.68 + 30 = 35.37 + 30 = 65.37

That's your answer.

Answer 3

Angle opposite the base being 32 degs, and knowing that the base angles of an isosceles triangle are equal:

180 - 2x = 32
2x = 148
x = 74

The base angles are each 74 degs.

At this point, you have a variety of options at your disposal to solve this problem. It sounds like this is a trig problem, so drop a vertical bisector from the non-base angle to form a right triangle. Using trig, you can quickly figure out the length of the hypotenuse.

Angles = 74, 90, 16
Length opposite 16 = 15 meters

cos(74) = 15/x

x = 54.42

Since this is an isosceles triangle, this length is the same for the third leg. Thus, the perimeter is:

30 + 2(54.42) = 138.84

Answer 4

Bisect the 32° angle to form two identical right triangles. Each triangle has base = 15m. The angle formed by the hypotenuse and base is 180-32/2-90 = 74°.

cos(74°) = 15/hypotenuse
hypotenuse = 15/cos(74°)
= 15/0.28
= 54.4m

perimeter of the isoceles triangle is 2*hypotenuse+30 = 138.8m

Answer 5

I can't draw you a picture so this is kinda hard to explain. Picture your triangle like ^ with your base on the bottom. Draw a line from the top to the middle of the base. It should intersect at 90 degrees. Since you have an angle and a right triangle, and you know that half of the base is 15 cm. You can use sin cos tan to find the other sides. Good Luck.

Answer 6

Drop a perpendicular from the 32 degree angle to the base. Now you have 2 triangles, each of which has angles of 16 degrees, 90 degrees and (by difference from 180) 74 degrees.

Cos 74 = .2756 = 15/x where x is one of the 2 equal legs of the isosceles triangle.

x = 54.427

Perimeter = 2 X 54.427 + 30 = 138.853

Answer 7

let ABC is isoceles triangle, where

base, BC = 30m

angle A = 32 degrees

now angle B + C = 180 - A = 180 - 32 = 148

since triangle is isosceles AB = AC

and
so B = C = 148/2 = 74 degrees

use sine rule

a/sinA = b/sinB = c/sinC

BC/Sin A = AC/sin B = AB/sin C

30/sin(32) = AC/sin(74)

AC = (30) * sin(74)/sin(32)

AC = 30(1.814) = 54.42 m

AB = 54.42

perimeter = AB + BC + AC

P = 2(54.42) + 30 = 138.84

Answer 8

because of the fact the suitable attitude is 32 deg, then 0.5 of is sixteen deg. The 0.5 triangle has base line 16m. one ninety deg attitude. one sixteen deg attitude and the 0.33 attitude is seventy 4 deg. sidelines are : a=b cos seventy 4= sixteen/b b= sixteen/cos seventy 4=fifty 8.04 m As a=b and the fringe is a+b+c= 30m + 2 x fifty 8.04= 148.08m

Answer 9

perimeter = 30 cm + 30 cm cos 16deg
= 30 cm [1 + 0.9613]
= 58.839 cm