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3 years ago

A,B & C form a triangle where BAC = 90°

Mathematics

1 answer:

kkurt [141]3 years ago

5 0

Answer:

12.2

Step-by-step explanation:

BC = √ABsq + √CAsq

BC = √11.9sq + √2.7sq

BC = √141.6 + √7.3

BC = √148.9

BC = 12.2

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Tasya [4]

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by comparison,

2x²-3=9-x²

3x²=12

x²=4

x= 2 or -2

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4 years ago

Please show working out, thanks!

sveticcg [70]

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4 0

3 years ago

Identify the property or rule 85+0=

frez [133]

IdentityProperty of Addition OR Zero Property

8 0

3 years ago

Read 2 more answers

URGENT HELP PLEASE!

Vaselesa [24]

Answer:

(a) $x=\dfrac{\pi}{8},\dfrac{3\pi}{8},\dfrac{9\pi}{8},\dfrac{11\pi}{8}$

(b) $x=\dfrac{3\pi}{2},\dfrac{\pi}{6},\dfrac{5\pi}{6}$

Step-by-step explanation:

It is given that $0\leq x\leq 2\pi$ .

(a)

$\sqrt{2}\sin 2x=1$

$\sin 2x=\dfrac{1}{\sqrt{2}}$

$\sin 2x=\dfrac{\pi}{4}$

$2x=\dfrac{\pi}{4},\dfrac{3\pi}{4},\dfrac{9\pi}{4},\dfrac{11\pi}{4}$ $[\because \sin x=\sin y\Rightarrow x=n\pi+(-1)^ny]$

$x=\dfrac{\pi}{8},\dfrac{3\pi}{8},\dfrac{9\pi}{8},\dfrac{11\pi}{8}$

(b)

$\csc^2 x-\csc x-2=0$

$\csc^2 x-2\csc x+\csc x-2=0$

$\csc x(\csc x-2)+1(\csc x-2)=0$

$(\csc x+1)(\csc x-2)=0$

$\csc x=-1\text{ or }\csc x=2$

$\sin x=-1\text{ or }\sin x=\dfrac{1}{2}$ $[\because \sin x=\dfrac{1}{\csc x}]$

$x=\dfrac{3\pi}{2}\text{ or }x=\dfrac{\pi}{6},\dfrac{5\pi}{6}$

Therefore, $x=\dfrac{3\pi}{2},\dfrac{\pi}{6},\dfrac{5\pi}{6}$ .

5 0

3 years ago

Find, correct to the nearest degree, the three angles of the triangle with the given vertices.

Leokris [45]

Answer:

angle CAB = 113.8 degree

angle ABC = 35.6 degree

angle BCA = 30.6 degree

Step-by-step explanation:

Given data:

A(1, 0, −1),

B(5, −3, 0),

C(1, 5, 2)

calculate the length of side by using the distance formula

AB = (5,-3,0) - (1,0,-1) = (4,-3,1)

AC= (1,5,2) - (1,0,-1) = (0,5,3)

|AB|

|AC| = $\sqrt {(0 + 5^2+3^2)} = \sqrt{34}$

From following formula, calculate the angle between the two side i.e Ab and AC

AB.AC = |AB|*|AC| cos ∠CAB

(4,-3,1).(0,5,3)

4*0 -3*5 +1*3

-12 =

cos ∠CAB = - 0.404

angle CAB = 113.8 degree

BA =B- A = (1,0,-1) - (5,-3,0) = (-4,3,-1)

BC = (1,5,2)-(5,-3,0) = (-4,8,2)

|BA| = \sqrt{(26)}

|BC| $= \sqrt {(4^2 + 8^2 + 2^2)} = \sqrt{(84)}$

BA.BC = |BA|*|BC|* cosABC

(-4,3,-1).(-4,8,2) = $\sqrt{(26)} * \sqrt{(84)} *cosABC$

16+24-2

cos ∠ABC = 0.813

angle ABC = 35.6 degree

we know sum of three angle in a traingle is 180 degree hence

sum of all three angle = 180

angle BCA + 35.6 + 113.8 = 180

angle BCA = 30.6 degree

7 0

4 years ago