Prove that sec^2θcosec*2θ-2-cot^2θ=tan2θ
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if ∡PQR = 82°, and the ray QS bisects it, it cuts ∡PQR into two equal halves, ∡PQS and ∡RQS, each of which is then 82/2, or 41°.
The length of q, to the nearest 10th of a centimeter is 7.6 cm.
Given in question,
In ΔOPQ,
o = 9.2 cm
p = 2.4 cm
∠Q = 37°
Cosine formula ⇒ cos θ =
Putting the values in equation,
cos 37 =
0.799 =
0.799*44.16 = 90.4 -
32.28 = 90.4 -
= 90.4 - 32.28
= 58.12
q = 7.6 cm (to nearest 10th)
Hence, length of q is 7.6 cm.
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