Question

In right angled ΔTSU, TS=5, ∠S=90°, SU=12 then find sin T, cos T, tan T. Similarly find sin U, cos U, tan U.

Answer 1

Hey buddy, your answer is in the picture

Answer 2

Answer:

$\rm \sin T = \dfrac{SU}{TU}=\dfrac{12}{13}.\\\cos T = \dfrac{TS}{TU}=\dfrac{5}{13}.\\\tan T = \dfrac{SU}{TS}=\dfrac{12}{5}.\\\\\\\sin U = \dfrac{TS}{TU}=\dfrac{5}{13}.\\\cos U = \dfrac{SU}{TU}=\dfrac{12}{13}.\\\tan U = \dfrac{TS}{SU}=\dfrac5{12}.$

Step-by-step explanation:

For a right-angled triangle, the trigonometric ratios are defined as

• The sine (sin) of an angle = $\rm \dfrac{Perpendicular}{Hypotenuse}.$

• The cosine (cos) of an angle = $\rm \dfrac{Base}{Hypotenuse}$

• The tangent (tan) of an angle = $\rm \dfrac{Perpendicular}{Base}$

For an angle in a right-angled triangle, the perpendicular is the side of the triangle which is opposite to that angle, the base is the side which is adjacent to that angle and the largest side of the triangle is hypotenuse.

Given:

In the given triangle TSU,

TS = 5.

SU = 12.

$\rm \angle S=90^\circ.$

According to Pythagoras theorem,

$\rm TU=\sqrt{SU^2+TS^2}=\sqrt{12^2+5^2}=\sqrt{144+25}=\sqrt{169}=13.$

For the angle T,

Perpendicular = SU.

Base = TS.

Hypotenuse = TU.

For the angle U,

Perpendicular = TS.

Base = SU.

Hypotenuse = TU.

Therefore,

$\rm \sin T = \dfrac{SU}{TU}=\dfrac{12}{13}.\\\cos T = \dfrac{TS}{TU}=\dfrac{5}{13}.\\\tan T = \dfrac{SU}{TS}=\dfrac{12}{5}.\\\\\\\sin U = \dfrac{TS}{TU}=\dfrac{5}{13}.\\\cos U = \dfrac{SU}{TU}=\dfrac{12}{13}.\\\tan U = \dfrac{TS}{SU}=\dfrac5{12}.$