Question

2. From the following figure, find the values of:
(i) cos B
(ii) tan C
(iii) sin’B + cos2B
(iv) sin B.cos C + cos B.sin C
​

Answer 1

Step-by-step explanation:

for angle B hypotenuse is BC=17 and base is AB=8

then cos B=8/17

and third side can be obtained as AC=

$\sqrt{17 {}^{2} - 8 {?}^{2} }$

=15

so sin B=15/17

I) therefore tanB=sin B/cos B

=(15/17)/(8/17)

=15/8

ii) also sin²B+cos²B is always =1

iii) cos C and sin C will be same as sin B and cos B respectively as they are opposite of each other to the right angle triangle

then sin C = cos B

= 8/17

and cos C = sin B

= 15/17

therefore sin B cos C+ cos B sin C= (8²/17)+ (15²/17)