22. The rod AC of a TV disc antenna is fixed
at right angles to the wall AB and a rod
CD is supporting the disc as shown in
Fig. 4. If AC=1.5 m long and CD = 3 m,
find (i) tan 0 (ii) sec 0 + cosec 0.
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If AC=1.5 m long and CD = 3 m, then,
(i) tan 0 = 0.579
(ii) sec 0 + cosec 0 = 3.158
Consider the figure while going through the following steps.
Given,
AC = 1.5 m
CD = 3 m
CD² = AD² + AC²
⇒ AD² = CD² - AC²
= 3² - 1.5² = 6.75
CD = 2.59 m
(i) tan 0
tan 0 = AC/AD
= 1.5/2.59
∴ tan 0 = 0.579
(ii) sec 0 + cosec 0
= CD/AD + CD/AC
= 3/2.59 + 3/1.5
∴ sec 0 + cosec 0 = 3.158
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Answer:
the answer is in the attachment
Step-by-step explanation:
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