Answer
Verified
412.5k+ views
Hint – In this question first of all use the Pythagoras theorem to get the value of side AC, that is ${\left( {{\text{AC}}} \right)^2} = {\left( {{\text{AB}}} \right)^2} + {\left( {{\text{BC}}} \right)^2}$, then use the concept that in triangle ABC when we consider angle A the perpendicular become BC and base become AB and hypotenuse remains same and we consider angle C the perpendicular become AB and base become BC and hypotenuse remains same. This will help get the values of the respective trigonometric angles.
Complete step-by-step answer:
The pictorial representation of the above problem is shown having right angle at B as shown in the figure.
It is given that AB = 24 cm, and BC = 7 cm.
Now apply Pythagoras theorem in triangle ABC we have,
$ \Rightarrow {\left( {{\text{Hypotenuse}}} \right)^2} = {\left( {{\text{perpendicular}}} \right)^2} + {\left( {{\text{base}}} \right)^2}$
$ \Rightarrow {\left( {{\text{AC}}} \right)^2} = {\left( {{\text{AB}}} \right)^2} + {\left( {{\text{BC}}} \right)^2}$
Now substitute the value we have,
$ \Rightarrow {\left( {{\text{AC}}} \right)^2} = {\left( {24} \right)^2} + {\left( 7 \right)^2} = 576 + 49 = 625 = {\left( {25} \right)^2}$
$ \Rightarrow AC = 25$ cm.
$\left( i \right)\sin A,\cos A$
Now as we know that sine is the ratio of perpendicular to hypotenuse.
So in triangle ABC when we consider angle A the perpendicular becomes BC and base becomes AB and hypotenuse remains the same.
$ \Rightarrow \sin A = \dfrac{{BC}}{{AC}} = \dfrac{7}{{25}}$
And cosine is the ratio of base to hypotenuse.
$ \Rightarrow \cos A = \dfrac{{AB}}{{AC}} = \dfrac{{24}}{{25}}$
$\left( {ii} \right)\sin C,\cos C$
Now as we know that sine is the ratio of perpendicular to hypotenuse.
So in triangle ABC when we consider angle C the perpendicular becomes AB and base becomes BC and hypotenuse remains the same.
$ \Rightarrow \sin C = \dfrac{{AB}}{{AC}} = \dfrac{{24}}{{25}}$
And cosine is the ratio of base to hypotenuse.
$ \Rightarrow \cos C = \dfrac{{BC}}{{AC}} = \dfrac{7}{{25}}$
So this is the required answer.
Note – It is advised to remember the direct formula for basic trigonometric ratios like $\sin \theta = \dfrac{P}{H},\cos \theta = \dfrac{B}{H},\tan \theta = \dfrac{P}{B}$ and $\cos ec\theta = \dfrac{H}{P},\sec \theta = \dfrac{H}{B},\cot \theta = \dfrac{B}{P}$. The trick point while considering different angles in a triangle is that the hypotenuse remains the same always however for the angles that are non-90 degrees the side which is exactly in front of it acts as perpendicular.
Complete step-by-step answer:
The pictorial representation of the above problem is shown having right angle at B as shown in the figure.
It is given that AB = 24 cm, and BC = 7 cm.
Now apply Pythagoras theorem in triangle ABC we have,
$ \Rightarrow {\left( {{\text{Hypotenuse}}} \right)^2} = {\left( {{\text{perpendicular}}} \right)^2} + {\left( {{\text{base}}} \right)^2}$
$ \Rightarrow {\left( {{\text{AC}}} \right)^2} = {\left( {{\text{AB}}} \right)^2} + {\left( {{\text{BC}}} \right)^2}$
Now substitute the value we have,
$ \Rightarrow {\left( {{\text{AC}}} \right)^2} = {\left( {24} \right)^2} + {\left( 7 \right)^2} = 576 + 49 = 625 = {\left( {25} \right)^2}$
$ \Rightarrow AC = 25$ cm.
$\left( i \right)\sin A,\cos A$
Now as we know that sine is the ratio of perpendicular to hypotenuse.
So in triangle ABC when we consider angle A the perpendicular becomes BC and base becomes AB and hypotenuse remains the same.
$ \Rightarrow \sin A = \dfrac{{BC}}{{AC}} = \dfrac{7}{{25}}$
And cosine is the ratio of base to hypotenuse.
$ \Rightarrow \cos A = \dfrac{{AB}}{{AC}} = \dfrac{{24}}{{25}}$
$\left( {ii} \right)\sin C,\cos C$
Now as we know that sine is the ratio of perpendicular to hypotenuse.
So in triangle ABC when we consider angle C the perpendicular becomes AB and base becomes BC and hypotenuse remains the same.
$ \Rightarrow \sin C = \dfrac{{AB}}{{AC}} = \dfrac{{24}}{{25}}$
And cosine is the ratio of base to hypotenuse.
$ \Rightarrow \cos C = \dfrac{{BC}}{{AC}} = \dfrac{7}{{25}}$
So this is the required answer.
Note – It is advised to remember the direct formula for basic trigonometric ratios like $\sin \theta = \dfrac{P}{H},\cos \theta = \dfrac{B}{H},\tan \theta = \dfrac{P}{B}$ and $\cos ec\theta = \dfrac{H}{P},\sec \theta = \dfrac{H}{B},\cot \theta = \dfrac{B}{P}$. The trick point while considering different angles in a triangle is that the hypotenuse remains the same always however for the angles that are non-90 degrees the side which is exactly in front of it acts as perpendicular.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
The branch of science which deals with nature and natural class 10 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Define absolute refractive index of a medium
Find out what do the algal bloom and redtides sign class 10 biology CBSE
Prove that the function fleft x right xn is continuous class 12 maths CBSE
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Write an application to the principal requesting five class 10 english CBSE
Difference Between Plant Cell and Animal Cell
a Tabulate the differences in the characteristics of class 12 chemistry CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
Discuss what these phrases mean to you A a yellow wood class 9 english CBSE
List some examples of Rabi and Kharif crops class 8 biology CBSE