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Mekhanik [1.2K]

3 years ago

7

Suppose a triangle has to size of lint 42 and 35 and the angles between these 2 sides is 120 which equated should you solve to f

ind the length of the 3rd side of a triangle

1 answer:

Brilliant_brown [7]3 years ago

5 0

Answer:

$7\sqrt{91}\ units$ or $66.78\ units$

Step-by-step explanation:

we know that

Applying the law of cosines

$c^{2}=a^{2}+b^{2}-2abcos(C)$

In this problem we have

$a=42\ units, b=35\ units, C=120\°$

c is the length of the third side

Substitute

$c^{2}=42^{2}+35^{2}-2(42)(35)cos(120\°)$

$c^{2}=1,764+1,225-2,940cos(120\°)$

$c^{2}=4,459$

$c=\sqrt{4,459}\ units$

$c=7\sqrt{91}\ units$ or $c=66.78\ units$

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Model 3 + (-4) on the number line

klemol [59]

So all your doing is going 3 jumps to the positive side and 4 jumps to the negative side.

4 0

3 years ago

Find, correct to the nearest degree, the three angles of the triangle with the given vertices. A(1, 0, âˆ’1), B(4, âˆ’3, 0), C(1

zavuch27 [327]

Answer:

Angle at vertex A : 59.2° B: 61.7° C : 59.1°

Step-by-step explanation:

1. First find the length of the vectors

AB = B-A = (4-1, 3-0, 0-1) = (3, 3, -1)

AC = C-A = (1-1, 4-0, 3-1) = (0, 4, 2)

BC = C-B = (1-4, 4-3, 3-0) = (-3, 1, 3)

2. Find magnitude of the vectors

|AB| = √(3^2+3^2+〖(-1)〗^2 )=√19

|AC| = √(0^2+4^2+2^2 )=2√5

|BC| = √(〖(-3)〗^2+1^2+3^2 )=√19

3. Find the angles between them

cos θ = (a.b)/(|a||b|) -----> θ = arc cos ((a.b)/(|a||b|))

AB, AC : θ =arc cos (AB.AC)/(|AB||AC|) = arc cos ( (3.0+3.4+ -1.2)/(√19 x 2√5)= 10/(2√95) ) =59.136, which is approximately 59.14°

AB, BC : θ =arc cos (AB.BC)/(|AB||BC|) = arc cos( (3.-3+3.1+ -1.3)/(√19 x √19)= (-9)/19 ) = 118.27°

Because of the direction of BC is pointing relative to AB this is the angle outside the triangle, and we should find the supplementary angle.

180-118.27 = 61.73°

If we used –(BC) = CB in the formula (just negate the numerator) we would have gotten the correct angle on first try.

The 3rd angle should be what’s left after subtracting from 180°;

180-61.73-59.14 = 59.13 °

You can confirm using the formula again :

AC, BC : θ =arc cos (AC.BC)/(|AC||BC|) = arc cos( (0.-3+4.1+ 2.3)/(√19 x 2√5)= 10/(2√95) = 59.13°

Rounding everything up so they add up to 180 degrees we have Angle at vertex A : 59.2° B: 61.7° C : 59.1°

7 0

3 years ago

Figure PQRS is a rectangle.

il63 [147K]

Answer:

4th option.

Step-by-step explanation:

PQ is not congruent to QR, instead it is PQ is perpendicular to QR.

6 0

3 years ago

If points S, O, and N are collinear, how many lines do they determine?

Gekata [30.6K]

It is only one line because a<span> set of points is collinear if they lie on a single straight line. This is very common to be applied in mathematics</span>

6 0

3 years ago

The sum of two terms of gp is 6 and that of first four terms is 15/2.Find the sum of first six terms.

Gnoma [55]

Given:

The sum of two terms of GP is 6 and that of first four terms is $\dfrac{15}{2}.$

To find:

The sum of first six terms.

Solution:

We have,

$S_2=6$

$S_4=\dfrac{15}{2}$

Sum of first n terms of a GP is

$S_n=\dfrac{a(1-r^n)}{1-r}$ ...(i)

Putting n=2, we get

$S_2=\dfrac{a(1-r^2)}{1-r}$

$6=\dfrac{a(1-r)(1+r)}{1-r}$

$6=a(1+r)$ ...(ii)

Putting n=4, we get

$S_4=\dfrac{a(1-r^4)}{1-r}$

$\dfrac{15}{2}=\dfrac{a(1-r^2)(1+r^2)}{1-r}$

$\dfrac{15}{2}=\dfrac{a(1+r)(1-r)(1+r^2)}{1-r}$

$\dfrac{15}{2}=6(1+r^2)$ (Using (ii))

Divide both sides by 6.

$\dfrac{15}{12}=(1+r^2)$

$\dfrac{5}{4}-1=r^2$

$\dfrac{5-4}{4}=r^2$

$\dfrac{1}{4}=r^2$

Taking square root on both sides, we get

$\pm \sqrt{\dfrac{1}{4}}=r$

$\pm \dfrac{1}{2}=r$

$\pm 0.5=r$

Case 1: If r is positive, then using (ii) we get

$6=a(1+0.5)$

$6=a(1.5)$

$\dfrac{6}{1.5}=a$

$4=a$

The sum of first 6 terms is

$S_6=\dfrac{4(1-(0.5)^6)}{(1-0.5)}$

$S_6=\dfrac{4(1-0.015625)}{0.5}$

$S_6=8(0.984375)$

$S_6=7.875$

Case 2: If r is negative, then using (ii) we get

$6=a(1-0.5)$

$6=a(0.5)$

$\dfrac{6}{0.5}=a$

$12=a$

The sum of first 6 terms is

$S_6=\dfrac{12(1-(-0.5)^6)}{(1+0.5)}$

$S_6=\dfrac{12(1-0.015625)}{1.5}$

$S_6=8(0.984375)$

$S_6=7.875$

Therefore, the sum of the first six terms is 7.875.

5 0

3 years ago