Answer:
Ques 11: 123.08 meters.
Ques 12: 269.4 meters.
Ques 13: 211.27 meters.
Explanation:
Ques 11: We have that the angle of elevation from the ship to the top of the lighthouse is 18° and the height of the lighthouse is 40 m.
It is required to find the distance of the ship from the shore, say 'x' m.
As, we have,
![\tan \theta=(opposite)/(adjacent)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/1y0f6m8zifxm4vkr7aqoz7bb0ivdpwua68.png)
⇒
![\tan 18=(40)/(x)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/5n48rbjgfyqnsvyh8omo9jdowiifhaiuxw.png)
⇒
![x=(40)/(\tan 18)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/96ck29jrqczug8s1v9xxw7q86vpdqd3c0z.png)
⇒
![x=(40)/(0.325)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/a5y15p03j2d8xpsd8km7xry81f29md4s0g.png)
⇒ x = 123.08 m.
Thus, the distance of the ship from the shore is 123.08 meters.
Ques 12: We have, length of the kite is 300 m and the angle made by the string is 64°.
It is required to find the height of the kite, say 'x' m.
As, we have,
![\sin\theta=(opposite)/(hypotenuse)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/cdz702icx3o2qdzpc6391gkv8s3pzf2za0.png)
⇒
![\sin 64=(x)/(300)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/vejns6ycfle2fjz3kfcuajm7a9lan813ka.png)
⇒
![x=300* \sin 64](https://img.qammunity.org/2020/formulas/mathematics/middle-school/qgsdvf59uvxai9xypptw8jn9tfcksknkc3.png)
⇒
![x=300* 0.898](https://img.qammunity.org/2020/formulas/mathematics/middle-school/seol2zd2272dj6qkfviuz6zlhkzvrqlco7.png)
⇒ x = 269.4 m.
Hence, the height of the kite is 269.4 meters.
Ques 13: We have, the wreckage is found at the angle of 12° and the diver is lowered down by 45 meters.
It is required to find the distance covered by the diver to reach the wreckage, say 'x' m.
As, we have,
![\tan \theta=(opposite)/(adjacent)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/1y0f6m8zifxm4vkr7aqoz7bb0ivdpwua68.png)
⇒
![\tan 12=(45)/(x)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/aos5gfm8trfuow7q6zozzlehegcq0drzl5.png)
⇒
![x=(45)/(\tan 12)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/zhwc8b3pgygehqu8eu7pp3xsk8b9fa2r26.png)
⇒
![x=(45)/(0.213)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/bjt3mdxdmedkcweykkmzilwxyqjj01vg9a.png)
⇒ x = 211.27 m.
Thus, the distance covered by the diver to reach the wreckage is 211.27 meters.