Which angles below are complementary?

Mathematics

1 answer:

tigry1 [53]2 years ago

6 0

9514 1404 393

Answer:

∠KXN and ∠QXT

Step-by-step explanation:

The measure of each angle is the difference of the scale values that the rays intercept. (The same scale needs to be used for each ray.) Of course, complementary angles have a sum of 90°.

Here, we'll refer to angle aXb as "ab".

NP = 95 -35 = 60

PQ = 35 -20 = 15 . . . not complementary to NP

KN = 165 -95 = 70

QT = 20 -0 = 20 . . . complementary to KN ⇒ your answer

JK = 180 -165 = 15

PQ = 35 -20 = 15 . . . not complementary to JK

JK = 15

NK = KN = 70 . . . not complementary to JK

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If sin A = 3/8, find the value of cosec A - sec A.

alexira [117]

Answer:

$\csc A - \sec A = \dfrac 83 + \dfrac{8}{\sqrt{55}}\\\\\csc A - \sec A = \dfrac 83 - \dfrac{8}{\sqrt{55}}$

Step by step explanation:

$\text{Given that,}\\\\~~~~~~\sin A = \dfrac 38 \\\\\implies \sin^2 A = \dfrac 9{64}\\\\\implies 1 - \cos^2 A = \dfrac{9}{64}\\\\\implies \cos ^2 A = 1 - \dfrac 9{64}\\\\\implies \cos^2 A = \dfrac{55}{64}\\\\\implies \cos A =\pm\sqrt{\dfrac{55}{64}}\\ \\\implies \cos A = \pm\dfrac{\sqrt{55}}8\\\\$

$\implies \dfrac 1{\cos A} = \pm\dfrac{8}{\sqrt{55}}$

$\text{Now,}\\\\\csc A - \sec A\\\\=\dfrac{1}{\sin A}- \dfrac{1}{\cos A}\\\\=\dfrac 83 -\left(\pm \dfrac 8{\sqrt {55}} \right)\\ \\\text{Hence,}\\\\\csc A - \sec A = \dfrac 83 + \dfrac{8}{\sqrt{55}}\\\\\csc A - \sec A = \dfrac 83 - \dfrac{8}{\sqrt{55}}$