Question

In the given figure, triangle ABC is right-angled at B. triangle BSC is right-angled at S and triangle BRS is right-angled at R. AB = 18 cm, BC = 7.5 cm, RS = 5 cm, RB = 6 cm, angle BSR = xº and angle SAB = yº. Find (1) tan xº (ii) sin yº (iii) cos y ​

Answer 1

Since triangle ABC is right-angled at B, we have:

$AB^2 + BC^2 = AC^2$

$18^2 + 7.5^2 = AC^2$

AC = √(18² + 7.5²)

= √(324 + 56.25)

= √(380.25)

= 19.5 cm

Since triangle BSC is right-angled at S, and triangle BRS is right-angled at R, we have:

BS² = BR² + RS²

$BS^2 = 6^2 + 5^2$

= 36 + 25 = 61

BS = √61

= 7.81 cm

Now, we have the lengths of AC and BS, and the angles xº and yº, so we can use trigonometry to find tan xº, sin yº, and cos yº.

Let's first find tan xº:

$tan x = RS/BR = 5/6 = 0.83$

Next, let's find sin yº:

$sin y = RS/BS = 5/7.81 = 0.64$

Finally, let's find cos yº:

$cos y = AC/BS = 19.5/7.81 = 2.51$

So,

(i) tan xº = 0.83

(ii) sin yº = 0.64

(iii) cos yº = 2.51

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