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3 years ago

Please help asap! One question!! Please I’m desperate!

Mathematics

2 answers:

Akimi4 [234]3 years ago

6 0

Answer:

A, B, C

Step-by-step explanation:

Use a^2 + b^2 = c^2 in each case.

3^2 + (sqrt(27))^2 = 6^2

9 + 27 = 36

36 = 36

A. does form a right triangle.

8^2 + 15^2 = 17^2

64 + 225 = 289

289 = 289

B. does form a right triangle.

5^2 + 5^2 = (sqrt(50))^2

25 + 25 = 50

50 = 50

C. does form a right triangle.

seraphim [82]3 years ago

4 0

Answer:

Step-by-step explanation:

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Use the diagram to complete the statement. Triangle J K L is shown. Angle K J L is a right angle. Angle J K L is 52 degrees and

zzz [600]

Answer:

$\bold{sin(38^\circ)=cos(52^\circ)}$

Step-by-step explanation:

Given that $\triangle KJL$ is a right angled triangle.

$\angle JKL = 52^\circ\\\angle KLJ = 38^\circ$

and

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Kindly refer to the attached image of $\triangle KJL$ in which all the given angles are shown.

To find:

sin(38°) = ?

a) cos(38°)

b) cos(52°)

c) tan(38°)

d) tan(52°)

Solution:

Let us use the trigonometric identities in the given $\triangle KJL$ .

We have to find the value of sin(38°).

We know that sine trigonometric identity is given as:

$sin\theta =\dfrac{Perpendicular}{Hypotenuse}$

$sin(\angle JLK) = \dfrac{JK}{KL}\\OR\\sin(38^\circ) = \dfrac{JK}{KL}$ ....... (1)

Now, let us find out the values of trigonometric functions given in options one by one:

$cos\theta =\dfrac{Base}{Hypotenuse}$

$cos(\angle JLK) = \dfrac{JL}{KL}\\OR\\cos(38^\circ) = \dfrac{JL}{KL}$ ....... (2)

By (1) and (2):

sin(38°) $\neq$ cos(38°).

$cos(\angle JKL) = \dfrac{JK}{KL}\\OR\\cos(52^\circ) = \dfrac{JK}{KL}$ ...... (3)

Comparing equations (1) and (3):

we get the both are same.

$\therefore \bold{sin(38^\circ)=cos(52^\circ)}$

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