Consider, in ΔRPQ,
RP = R (Radius of larger circle)
PQ = r (radius of smaller circle)
We have to find, RQ, by Pythagoras theorem,
RP² = PQ²+RQ²
R² = r²+RQ²
RQ² = R²-r²
RQ = √(R²-r²
Now, as RQ & QS both are tangents of the smaller circle, their lengths must be equal. so, RS = 2 × RQ
RS = 2√(R²-r²)
In the parallelogram ABCD, join BD.
Consider the triangle Δ ABD.
It is given that AB > AD.
Since, in a triangle, angle opposite to longer side is larger, we have,
∠ ADB > ∠ ABD. --- (1)
Also, AB || DC and BD is a transversal.
Therefore,
∠ ABD = ∠ BDC
Substitute in (1), we get,
∠ ADB > ∠ BDC.
Answer:
5^6=15625
Step-by-step explanation: the little thing on the right of log to the power of the left of = is equal to the big numbers right of log
Answer:
a hope help you stay happy and safe
