Question

From the following figure, find the values of
(1) sin B
(2) tan c
(3) sec^2 B - tan^2 B (iv) sin^2 C + cos^2 C​

Answer 1

Answer:

1.sinb=12/13

2.12/16 or3/4

3.1

4.1

Answer 2

Answer:

here,

in triangle ABD,

AB=13

BD=5

By pythagoras theorem,

$ad = \sqrt{ab {}^{2} - {bd}^{2} } = \sqrt{ {13}^{2} - {5}^{2} } = \sqrt{169 - 25} = \sqrt{144} = 12$

In triangle ADC,

AD=12

CD=16

By pythagoras theorem

$ac = \sqrt{ {ad}^{2} + {cd}^{2} } = \sqrt{12 {}^{2} + {16}^{2} } = \sqrt{144 + 256} = \sqrt{400} = 20$

☆1☆

$\sin(b) = \frac{ad}{ab} = \frac{12}{13}$

☆2☆

$\tan(c) = \frac{ad}{cd} = \frac{12}{16} = \frac{3}{4}$

☆3☆

${ \sec }^{2} b - { \tan }^{2} b \\ = (\frac{ab}{bd} ) ^{2} - ( \frac{ad}{bd} ) {}^{2} \\ = (\frac{13}{5} ) {}^{2} - ( \frac{12}{5} ) ^{2} \\ = \frac{169}{25} - \frac{144}{25} \\ = \frac{169 - 144}{25} \\ = \frac{25}{25} \\ = 1$

☆4☆

$\sin^{2} c + \cos^{2} c \\ = ( \frac{12}{20} ) {}^{2} + (\frac{16}{20} ) {}^{2} \\ = \frac{144}{400} + \frac{256}{400} \\ = \frac{144 + 256}{400} \\ = \frac{400}{400} \\ = 1$