Question

please answer the following ​

Answer 1

Step-by-step explanation:

19. ∠B = 90°, AB = 8cm

$\sin{A} = \frac{3}{5} = \frac{p}{h}$

Using Pythagoras Theorem,

${b}^{2} = {h}^{2} - {p}^{2}$

${b}^{2} = 25 - 9$

${b}^{2} = 16$

$b = 4$

Now let the sides be b=4x, p=3x and h=5x

Given that b = 8cm

. 4x = 8cm

. x = 2cm

∴p = 3x = 3 × 2cm = 6cm

∴p = 3x = 3 × 2cm = 6cm. h = 5x = 5 × 2cm = 10cm

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20. ∠B = ∠D = 90°, AE = 26cm, CE = 10cm

$\sin\alpha = \frac{3}{5} , \cos\beta = \frac{5}{13}$

In ΔCDE, using Pythagoras Theorem,

b² = h² - p²

b² = 5² - 3²

b² = 25 - 9

b² = 16

b = 4

Now, in ΔCDE, let b = 4x and h = 5x

Given that h = 10cm

. 5x = 10cm

. x = 2cm

∴b = 4x = 4 × 2cm = 8cm

Now, in ΔABE, using Pythagoras Theorem,

p² = h² - b²

p² = 13² - 5²

p² = 169 - 25

p² = 144

p = 12

Now, in ΔABE, let b = 5x and h = 13x

Given that h = 26cm

. 13x = 26cm

. x = 2cm

∴b = 5x = 5 × 2cm = 10cm

∴BD = BE + ED = 10cm + 8cm = 18cm

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21. ∠ABD = ∠ BDC = 90°, CD = 5cm, BC = 13cm, AB = 9cm

In ΔBDC, Using Pythagoras Theorem,

BD² = BC² - CD²

b² = (13cm)² - (5cm)²

b² = 169cm² - 25cm²

b² = 144cm²

b = 12cm

$∴ \bold{\tan \alpha = \frac{5}{12}}$

Now, in ΔABD, Using Pythagoras Theorem,

AD² = AB² + BD²

h² = (9cm)² + (12cm)²

h² = 81cm² + 144cm²

h² = 225cm²

h = 15cm

$\bold{ \sin\beta = \frac{9}{15} = \frac{3}{5} }$

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22. ∠BAC = ∠ACD = 90°, AB = 10cm, BC = 26cm, CD = 28cm

In ΔBAC, using Pythagoras Theorem,

b² = h² - p²

b² = (26cm)² - (10cm)²

b² = 676cm² - 100cm²

b² = 576 cm²

b = 24cm

$\bold{\sin\alpha = \frac{10}{26} = \frac{5}{13} }$

$\bold{ \cos \alpha = \frac{24}{26} = \frac{12}{13} }$

Now, drawing line BE||AC, AC = BE = 24cm, AB = CE = 10cm, DE = 28cm - 10cm = 18cm

In ΔADE, using Pythagoras Theorem,

AD² = DE² + AE²

h² = (18cm)² + (24cm)²

h² = 324cm² + 576cm²

h² = 900cm²

h = 30cm

$\bold{ \tan\beta = \frac{24}{18} = \frac{4}{3} }$

$\bold{ \csc \beta = \frac{30}{24} = \frac{5}{4} }$

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