Math Problem?

Question

Find the area of triangle ABC where
angle ABC = 35 degrees
angle BCA = 110 degrees
angle CAB = 35 degrees
AND, length BC = 4 cm

Answer 1

if BC = 4 cm then AC = 4 cm

area = (1/2) * BC * AC * sin(ACB) --->same as (1/2)*base*height
area = (1/2) * 4* 4* sin(110)

area = 7,517541 cm^2

Answer 2

the first thing you need to do is to draw out the triangle. what you will get is an isosceles triangle.
since BC is 4cm, the isosceles property of the triangle will make AC 4 cm as well.

once you have obtained this you can implement the formula:
1/2 x a x b x sinc to get the area of the triangle.
a and b are any 2 sides of the triangle while c is the angle between the two sides.
from your drawing you can deduce that a = b = 4cm and the angle between them is 110 degrees.

therefore, the area is = 1/2 x 4 x 4 x sin 110
= 7.52cm^2 (3sf)

hope this helped. drawing the diagram out and knowing what formulas are at your disposal is really very helpful. so remember to draw and start getting well acquainted with your formulas!
have fun with math

Answer 3

Sketch the triangle and identify the angles/length. You'll find it's an Isosceles triangle with 2 equal sides.

Find the perpendicular height by bisecting angle BCA and drawing a line to the middle of side AB from the angle BCA. Lets call this D:

............C
......./....|.....\
.../........|........\
.A---------D---------B

Where ADC is 90 degrees.
So ACD is 55 degrees (half 110)
CAB =CAD = 35 degrees
and length AC = 4cm

Find perpendicular height DC using
sin CAD = DC/AC (opp/Hyp)
sin 35 = DC/4
4xsin 35 = DC
4x0.574 = DC
DC = 2.294cm

Find Base length:
Use Pythag to find length AD:
AC^2 = AD^2 + DC^2 (^2 means squared)
4^2 = AD^2 + 2.294^2
16 = AD^2 + 5.2638
so
AD^2= 16 - 5.2638
AD^2= 10.736
square rooting gives:
AD = 3.2766cm

so base length AB = 2 x 3.2766
AB = 6.5532cm

Now Area of triangle is 1/2 x base x perp. height

Area = 1/2 x 6.5532 x 2.294
= 7.51754cm^2
Area of triangle ABC = 7.52cm^2 (to 2 dp)
thats decimal places!

Answer 4

ABC = 35°
BCA = 110°
CAB = 35°

BC = 4 cm

Since BC is 4cm, then AC is 4cm, because their angles are the same

AC = 4cm
BC = 4cm
AB = Unknown

Lets call these sides

AC = side b
BC = side a
AB = side c

so now we have

side a = 4
side b = 4
side c = unknown

Using c^2 = a^2 + b^2 - 2ab(cosC)

c^2 = 4^2 + 4^2 - 2(4 * 4 * cos(110))
c^2 = 16 + 16 - 32cos(110)
c^2 = 32 - 32cos(110)
c^2 = 32(1 - cos(110))
c = sqrt(32(1 - cos(110)))
c = sqrt(16(2 - 2cos(110)))
c = 4sqrt(2 - 2cos(110))
c = about 6.5532163543119...
c = about 6.55

now using

Area = ac * sin(A)/2

Area = (4 * 4sqrt(2 - 2cos(110))) * ((sin(35))/2)
Area = (16sqrt(2 - 2cos(110)) * sin(35))/2
Area = (8sin(35))(sqrt(2 - 2cos(110)))
Area = about 7.517540966287267...
Area = about 7.52 cm^2

Even if you used Area = b * sqrt(4a^2 - b^2)/4

You would still get

ANS : about 7.52cm^2

Answer 5

Since, from your information, angle ABC is congruent to angle CAB, they have the same measures, and the triangle is isosceles, with AC = BC.
Therefore, AC = BC = 4 cm

Area = 1/2 (AC)(BC) sin (BCA)
Area = 1/2(4 cm)(4 cm) sin (110º)

Therefore,
Area = 8 sin (110º) cm² ≈ 7.5175 cm²

^_^

Answer 6

Draw the bisector of the obtuse angle. You now have two right triangles, and you've been given the length of the hypotenuse. Multiply by sine of the given acute angle to get the length of the bisector you drew, which is also the altitude of the original triangle. Multiply the length of the hypotenuse by cosine of the same given angle to get the length of the other leg of the right triangle you constructed, and double it to get the length of the base of the original triangle. Now you have the base and height of the original triangle, and area equals one half base times height.

Answer 7

7.5cm^2
Since it's an isosceles triangle with the pair of equal
angles 35d you can drop a perpendicular bisector
from C to AB to a pt D and you have 2 rt triangles of equal area. Note that sin(35)=CD/BC=CD/4.
So CD=4sin(35)=2.29.
Also cos(35)=BD/4, so BD=4cos35=3.28.
The area of ABC is 2Xarea BDA and
area BDA is 2.29(3.28)/2 so
area ABC is 2.29(3.28)=7.5cm

Answer 8

Remember, Area = (a^2 x sinB x sinC)/(2 x sinA)

So, Area = (BC^2 x sinABC x sin BCA)/(2 x sinCAB)
= 16sin35sin110/(2 xsin35)
= 8sin110
= 8 x 0.9397
= 7.517 sq.unit

Answer 9

Area = 7.5175
AB = 6.5532
AC = 4

Answer 10

The answer is : S = 8