Question

★ Trigonometry ★

1: In ∆ ABC , right-angled at B , AB = 24 cm , BC = 7 cm. Determine :
• sin A
• cos A
• sin C
• cos C

2: Given sec$\theta$ = $\dfrac{13}{12}$ , calculate all other trigonometric ratios.

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Answer 1

Question 1.

1: In ∆ ABC , right-angled at B , AB = 24 cm , BC = 7 cm. Determine :

• sin A
• cos A
• sin C
• cos C

Required Answer:-

Given:-

In ∆ ABC , B is at right angle.

• AB = 24 cm
• BC = 7 cm

To Find:-

To determine:-

• sin A
• cos A
• sin C
• cos C

Solution:-

◇Kindly see the 1st attachment

We know that, According to Pythagoras Theorem :-

$\\ \underline{ \boxed{ \pmb{Hypotenuse {}^{2} = Perpendicular {}^{2} + Base {}^{2} }}}$

Hence,

$\\ \sf{ \bold{ {AC}^{2} = AB {}^{2} + {BC}^{2} }}$

$\\ \sf{ \implies{ {AC}^{2} = {24}^{2} + {7}^{2} }}$

$\\ \sf{ \implies{ {AC}^{2} = 576 + 49}}$

$\\ \sf{ \implies{AC = \sqrt{625} }}$

$\\ \sf{ \therefore{AC = \bold{25cm}}}$

Therefore,

As we know that:-

$\\ \sf{ \bold{sin \theta = \frac{Opposite \: side}{hypotenuse}}}$

$\\\sf{ \rightarrow{ sinA = \frac{BC}{AC}}}$

$\sf{ \therefore{ sinA = \bold{ \red{\frac{7}{25}}} }}$

$\\ \sf{ \bold{cos \theta = \frac{Adjacent \: side}{Hypotenuse}}}$

$\\ \sf{ \rightarrow{cosA = \frac{AB}{AC} }}$

$\\\sf{ \therefore{cosA= \bold{ \red{ \frac{24}{25} }}}}$

Similarly,

$\\ \sf{ \rightarrow{ sinC = \frac{AB}{AC}}}$

$\\\sf{ \therefore{ sinC = \bold{ \red{\frac{24}{25}}} }}$

And,

$\\\sf{ \rightarrow{cosC= \frac{BC}{AC} }}$

$\\\sf{ \therefore{cosC= \bold{ \red{ \frac{7}{25} }}}}$

Question 2.

$\sf{2: Given \: sec\theta = \dfrac{13}{12} }$

Calculate all other trigonometric ratios.

Required Answer:-

Given:-

$\sf{ sec\theta = \dfrac{13}{12} }$

To Find :-

• Calculate all other trigonometric ratios.

Solution:-

◇ Kindly see the 2nd Attachment.

Let △ABC be right angled triangle (right angled at B)

As we know that:-

$\\ \underline{ \boxed{ \pmb{sec \theta = \frac{hypotenuse}{Adjacent \: side}}}}$

As we were given, we can write,

$\\ \sf{ \bold{sec \theta = \frac{13}{12} = \frac{AC}{AB}}}$

Now, according to Pythagoras Theorem :-

$\\ \underline{ \boxed{ \pmb{Hypotenuse {}^{2} = Perpendicular {}^{2} + Base {}^{2} }}}$

Hence,

$\: \\ \sf{ \bold{ {AC}^{2} = AB {}^{2} + {BC}^{2} }}$

$\\ \sf{ \implies{BC {}^{2} = {AC}^{2} - {AB}^{2} }}$

$\\ \sf{ \implies{ {BC}^{2} = {13}^{2} - {12}^{2} }}$

$\\ \sf{ \implies{ {BC}^{2} = 169 - 144}}$

$\\ \sf{ \implies{BC = \sqrt{25}}}$

$\\ \sf{ \therefore{BC= \bold{5cm}}}$

Now,

$\\ \sf{ \bold{sin \theta = \frac{Opposite \: Side}{Hypotenuse} }}$

$\\ \sf{ \rightarrow{sin \theta = \frac{BC}{AC} = \red{ \frac{5}{13}}}}$

$\\ \sf{ \bold{cos \theta = \frac{Adjacent \: Side}{Hypotenuse} }}$

$\\ \sf{ \rightarrow{cos\theta = \frac{AB}{AC} = \red{ \frac{12}{13}}}}$

$\\ \sf{ \bold{tan \theta = \frac{Opposite \: Side}{Adjacent \: side} }}$

$\\ \sf{ \rightarrow{tan\theta = \frac{BC}{AB} = \red{ \frac{5}{12}}}}$

$\\ \sf{ \bold{cot\theta = \frac{Adjacent \: Side}{Opposite\: side} }}$

$\\ \sf{ \rightarrow{cot\theta = \frac{AB}{BC} = \red{ \frac{12}{5}}}}$

$\\ \sf{ \bold{cosec \theta = \frac{Hypotenuse}{Opposite \: Side} }}$

$\\ \sf{ \rightarrow{cosec \theta = \frac{AC}{BC} = \red{ \frac{13}{5}}}}$

More Information:-

• sin θ = Opposite Side/Hypotenuse

• cos θ = Adjacent Side/Hypotenuse

• tan θ = Opposite Side/Adjacent Side

• sec θ = Hypotenuse/Adjacent Side

• cosec θ = Hypotenuse/Opposite Side

• cot θ = Adjacent Side/Opposite Side

Answer 2

Sum no.1

Given:-

Triangle ABC, right angled at B=90°

AB=24 cm

BC=7 cm

To find:-

Determine:

sin A

cos A

sin C

cos C

Solution:-

In a given triangle ABC, right angle at B=90°.

Given that,

AB=24 cm

BC=7 cm

According to pythagoras theorem,

In a right angled triangle,the squares of the hypotenuse side is equal to the sum of the squares of the other two sides.

By applying pythagoras theorem,

$\sf\fbox {AC²=AB²+BC²}$

Substituting the values,we get,

$\sf{AC²=(24)²+7²} \\ </p><p> \sf{AC²=(576+49)} \\ </p><p> \sf{AC²=625 cm²} \\ </p><p> \sf{AC= \sqrt{625} =25}</p><p></p><p>$

Therefore,AC=25 cm.

✠ Solution Ꭵ

To find Sin (A), Cos (A)

We know that,

Sin function is the equal to the ratio of the opposite side to the hypotenuse side.

According to question,

$\sf{{sin A = \frac{opposite \: side}{hypotenus} }}$

Substituting the values,

$\sf{sin \: A = \frac{ab}{ac} = \frac{24}{25} }$

Cos function is equal to the ratio of the length of adjacent side to the hypotenuse side.

Hence it becames,

$\sf{sin \: A = \frac{oppposite \: side}{hypotenuse} = \frac{24}{25} }$

✠ Solution ᎥᎥ

To find Sin (C), Cos (C)

$\sf{ \sin(c) = \frac{ab}{ac} = \frac{24}{25} }$

$\sf{\cos(c) = \frac{bc}{ac} = \frac{7}{25} }$

Sum no.2

$\frak{ \sec(a) = \frac{13}{12} = \frac{h}{b} }$

By Pythagoras theorem,

$\sf{ {bc}^{2} = {ac}^{2} + {ab}^{2} } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: = (13 {)}^{2} - (12 {)}^{2} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 169 - 144 = 25$

also,

$\sf{sin \:A = \frac{5}{13} } \\ \sf{cos \: A = \frac{12}{13} } \\ \sf{tan \: A = \frac{5}{12} } \\ \sf{cot \: A = \frac{12}{5} } \\ \sf {cosec \:A = \frac{13}{5} }$