Question

In angle ABC is a right angle at angle A. Find ten, sir and cos of angle B and C in each of the following cases
(I) Ab = 7cm, AC =24cm
(ii) AB = 12cm, AC =9cm

Answer 1

Answer:

triangle the additive inverse of the most important part of oue

Answer 2

Answer:

1. tan∠B =3.428 ; sin∠B=0.96 ; cos∠B= 0.28.                                          tan∠C= 0.29166 ; sin∠C= 0.28 ; cos∠C=0.96
2. tan∠B= 0.75 ; sin∠B= 0.6 ; cos∠B= 0.8                                                              tan∠C= 1.33 ; sin∠C= 0.8 ; cos∠C= 0.6

Step-by-step explanation:

(1) We know, for a right angled triangle, If A=90°, then BC is the hypotenuse.        

According to Pithagorus Theoram,

          Hypotenuse²= Side² + Side²

        ⇒  BC² = AC² + AB²

        ⇒  BC=√(24² + 7²)

         ∴  BC= 25

Furthermore, In case of any angle ∠α , sin∠α=$\frac{height}{hypotenuse}$

                                                                 cos∠α=$\frac{Base }{hypotenuse}$

                                                     and      tan∠α=$\frac{Height}{Base}$

   Point to be noted that: the arm adjacent to ∠α is base and the other one is height.

If we consider ∠B as ∠α then , AB is Base and AC is height.

    Therefore,    sin∠B=$\frac{Height}{Hypotenuse}$ =$\frac{AC}{BC}$ =$\frac{24}{25}$ = 0.96

                          cos∠B=$\frac{Base }{Hypotenuse}$ =$\frac{AB}{BC}$ =$\frac{7}{25}$ = 0.28

                          tan∠B=$\frac{Height}{Base}$ =$\frac{AC}{AB}$ = $\frac{24}{7}$ = 3.428

Again If we consider ∠C as ∠α then , AC is Base and AB is Height.

Therefore,    sin∠C=$\frac{Height}{Hypotenuse}$ = $\frac{AB}{BC} = \frac{7}{25} = 0.28$

                          cos∠C=$\frac{Base }{Hypotenuse}$ = $\frac{AC}{BC} = \frac{24}{25} = 0.96$

                          tan∠C=$\frac{Height}{Base}$ = $\frac{AB}{AC} = \frac{7}{24} = 0.29166$

(2) Same process for No. 2. Just change the value.