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2 years ago

What is the tangent ratio of angle x°

Mathematics

2 answers:

enot [183]2 years ago

8 0

Tan=opposite/adjacent. Opposite=6, adjacent=8. Tan=6/8. We can divide both terms by 2 to simplify to tan=3/4

almond37 [142]2 years ago

7 0

Answer:

tan x = 3/4

Step-by-step explanation:

The hypotenuse of the triangle is 10 ft.

For angle x, the opposite leg is 6 ft, and the adjacent leg is 8 ft.

tan A = opp/adj

tan x = 6 ft / 8 ft

tan x = 3/4

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Given: KL ║ NM , LM = 45, m∠M = 50° KN ⊥ NM , NL ⊥ LM Find: KN and KL

Mice21 [21]

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Step-by-step explanation:

Given:

KL ║ NM ,

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1. Consider triangle NLM. This is a right triangle, because NL ⊥ LM. In this triangle,

LM = 45

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So,

$\tan \angle M=\dfrac{\text{opposite leg}}{\text{adjacent leg}}=\dfrac{NL}{LM}=\dfrac{NL}{45}\\ \\NL=45\tan 50^{\circ}$

Also

$m\angle LNM=90^{\circ}-50^{\circ}=40^{\circ}$ (angles LNM and M are complementary).

2. Consider triangle NKL. This is a right triangle, because KN ⊥ NM . In this triangle,

$NL=45\tan 50^{\circ}$

$m\angle KLN=m\angle LNM=40^{\circ}$ (alternate interior angles)

$m\angle KNL=90^{\circ}-40^{\circ}=50^{\circ}$ (angles KNL and KLN are complementary).

So,

$\sin \angle KNL=\dfrac{\text{opposite leg}}{\text{hypotenuse}}=\dfrac{KL}{LN}=\dfrac{KL}{45\tan 50^{\circ}}\\ \\KL=45\tan 50^{\circ}\sin 50^{\circ}\approx 41.08$

and

$\cos \angle KNL=\dfrac{\text{adjacent leg}}{\text{hypotenuse}}=\dfrac{KN}{LN}=\dfrac{KN}{45\tan 50^{\circ}}\\ \\KN=45\tan 50^{\circ}\cos 50^{\circ}=45\sin 50^{\circ}\approx 34.47$

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What is the tangent ratio of angle x°​

What is the tangent ratio of angle x°