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Text Solution

`10sqrt3`m15 m`15sqrt3`m20 m

Answer :

DSolution :

AB be the height of the building. <br> `thereforeAB=5` m(given) <br> Let PQ be the height of the tower. <br> In `DeltaABP`, <br> `tan30^(@)=(AB)/(AP)rArr(1)/(sqrt3)=(5)/
(AP)`. <br> `therefore=5sqrt3 rArrthereforeBR=5sqrt3` m <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/PS_MATH_X_C08_E06_017_S01.png" width="80%"> <br> In `DeltaBQR, tan60^(@)=(QR)/(BR)` <br> `sqrt3=(QR)/(5sqrt3)rArrQR=15`m <br> From the figure, PQ = PR + QR <br> =5 + 15 = 20 m. **Application of trignometry in real life.**

**Explain the angle of elevation & depression with the help of diagram.**

**A circus artist is climbing a 20 m long rope ; which is tightly stretched and tied from the top of a vertical pole to the ground . Find the height of the pole if the angle made by the rope with the ground level is `30^@`**

**From the top of the hill; the angle of depressions of two consecutive kilometre stones due east are found to be `30^@` and `45^@` . Find the height of the hill.**

**The shadow of a flag staff is 3 times as long as shadow of the flag-staff when the sunrays meet the ground at an angle of `60^@`. Find the angle between the sun rays and the ground at the time of longer shadow.**

**The angle of elevation of a cloud from a point 60m above a lake is `30^@` and the angle of depression of the reflection of cloud in the lake is `60^@`. Find the height of the cloud.**

**From the top of a building 15m high the angle of elevation of the top of tower is found to be `30^@`. From the bottom of same building ; the angle of elevation of the top of the tower is found to be `60^@`. Find the height of the tower and the distance between tower and building .**

**A man on a cliff observes a boat at an angle of depression of `30^@` which is approaching the shore to the point immediately beneath the observer with a uniform speed . Six minutes later ; the angle of depressions of the boat is found to be `60^@`. Find the time taken by boat to reach the shore.**

**The elevation of a tower at a Station A due north of it is `alpha` and at a station B due west of A is `beta`. Prove that the height of the tower is `(ABsinalphasinbeta)/sqrt(sin^alpha-sin^beta)`**